NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry

NCERT Solutions for Class 10 Maths Chapter 7, "Coordinate Geometry" covers the following four exercises:

Exercise 7.1 - In this exercise, students are introduced to the concept of the Cartesian Plane, plotting points on the plane, and finding the coordinates of a point. Students learn how to identify the quadrants of the plane and the axes (x-axis and y-axis). The exercise contains questions related to finding the coordinates of a point, plotting a point on the plane, and identifying the quadrant in which a point lies.

Exercise 7.2 - In this exercise, students learn to find the distance between two points on a Cartesian Plane using the distance formula. They are also taught how to apply the midpoint formula to find the midpoint of a line segment. This exercise contains questions related to finding the distance between two points, finding the midpoint of a line segment, and applying the distance and midpoint formulas to solve problems.

Exercise 7.3 - In this exercise, students learn about the slope of a line, how to find the slope of a line, and how to determine if two lines are parallel, perpendicular or neither. The exercise contains questions related to finding the slope of a line, determining if two lines are parallel, perpendicular or neither, and finding the equations of lines given their slopes and a point on the line.

Exercise 7.4 - In this exercise, students learn to find the equation of a line given its slope and a point on the line. They also learn how to find the equation of a line parallel or perpendicular to a given line passing through a given point. This exercise contains questions related to finding the equation of a line given its slope and a point, finding the equation of a line parallel or perpendicular to a given line, and using the point-slope form to find the equation of a line.

Access NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry

Exercise (7.1)

1. Find the distance between the following pairs of points: 

(i) $\left( 2,3 \right),\left( 4,1 \right)$

Ans:

Given that,

Let the points be $\left( 2,3 \right)$ and $\left( 4,1 \right)$

To find the distance between the points $\left( 2,3 \right),\left( 4,1 \right)$.

Distance between two points is given by the Distance formula $=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

Here, ${{x}_{1}}=2$

 ${{x}_{2}}=4$

${{y}_{1}}=3$

${{y}_{2}}=1$  

Thus, the distance between $\left( 2,3 \right)$ and $\left( 4,1 \right)$ is given by,

$d=\sqrt{{{\left( 2-4 \right)}^{2}}+{{\left( 3-1 \right)}^{2}}}$

$=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 2 \right)}^{2}}}$

$=\sqrt{4+4}$

$=\sqrt{8}$

$=2\sqrt{2}$

$\therefore $The distance between $\left( 2,3 \right)$ and $\left( 4,1 \right)$ is $2\sqrt{2}$ units.

(ii) $\left( -5,7 \right),\left( -1,3 \right)$

Ans:

Given that,

Let the points be $\left( -5,7 \right)$ and $\left( -1,3 \right)$

To find the distance between the points $\left( -5,7 \right),\left( -1,3 \right)$.

Distance between two points is given by the Distance formula $=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

Here, ${{x}_{1}}=-5$

${{x}_{2}}=-1$

${{y}_{1}}=7$

${{y}_{2}}=3$  

Thus, the distance between $\left( -5,7 \right)$ and $\left( -1,3 \right)$ is given by,

$d=\sqrt{{{\left( -5-\left( -1 \right) \right)}^{2}}+{{\left( 7-3 \right)}^{2}}}$

$=\sqrt{{{\left( -4 \right)}^{2}}+{{\left( 4 \right)}^{2}}}$

$=\sqrt{16+16}$

$=\sqrt{32}$

$=4\sqrt{2}$

$\therefore $The distance between $\left( -5,7 \right)$ and $\left( -1,3 \right)$ is $4\sqrt{2}$ units.

(iii) $\left( a,b \right),\left( -a,-b \right)$

Ans:

Given that,

Let the points be $\left( a,b \right)$ and $\left( -a,-b \right)$

To find the distance between the points $\left( a,b \right),\left( -a,-b \right)$.

Distance between two points is given by the Distance formula $=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

Here, ${{x}_{1}}=a$

${{x}_{2}}=-a$

${{y}_{1}}=b$

${{y}_{2}}=-b$  

Thus, the distance between $\left( a,b \right)$ and $\left( -a,-b \right)$ is given by,

$d=\sqrt{{{\left( a-\left( -a \right) \right)}^{2}}+{{\left( b-\left( -b \right) \right)}^{2}}}$

$=\sqrt{{{\left( 2a \right)}^{2}}+{{\left( 2b \right)}^{2}}}$

$=\sqrt{4{{a}^{2}}+4{{b}^{2}}}$

$=\sqrt{4}\sqrt{{{a}^{2}}+{{b}^{2}}}$

$=2\sqrt{{{a}^{2}}+{{b}^{2}}}$

$\therefore $The distance between $\left( a,b \right)$ and $\left( -a,-b \right)$ is $2\sqrt{{{a}^{2}}+{{b}^{2}}}$ units.

2. Find the distance between the points $\left( 0,0 \right)$ and $\left( 36,15 \right)$. Can you now find the distance between the two towns $A$ and $B$ discussed in Section 7.2?

Ans:

Given that,

Let the points be $\left( 0,0 \right)$ and $\left( 36,15 \right)$

To find the distance between the points $\left( 0,0 \right),\left( 36,15 \right)$.

Distance between two points is given by the Distance formula $=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

Here, ${{x}_{1}}=0$

${{x}_{2}}=36$

${{y}_{1}}=0$

${{y}_{2}}=15$  

Thus, the distance between $\left( 0,0 \right)$ and $\left( 36,15 \right)$ is given by,

$d=\sqrt{{{\left( 0-36 \right)}^{2}}+{{\left( 0-15 \right)}^{2}}}$

$=\sqrt{{{\left( -36 \right)}^{2}}+{{\left( -15 \right)}^{2}}}$

$=\sqrt{1296+225}$

$=\sqrt{1521}$

$=39$

Yes, it is possible to find the distance between the given towns $A$ and $B$. The positions of this town are $A\left( 0,0 \right)$ and $B\left( 36,15 \right)$. And it can be calculated as above.

$\therefore $The distance between $A\left( 0,0 \right)$ and $B\left( 36,15 \right)$ is $39$ km.

3. Determine if the points $\left( 1,5 \right),\left( 2,3 \right)$ and $\left( -2,-11 \right)$ are collinear.

Ans:

Given that,

Let the three points be $\left( 1,5 \right),\left( 2,3 \right)$ and $\left( -2,-11 \right)$

To determine if the given points are collinear

Let $A\left( 1,5 \right),B\left( 2,3 \right),C\left( -2,-11 \right)$ be the vertices of the given triangle.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $A\left( 1,5 \right)$ and $B\left( 2,3 \right)$

${{x}_{1}}=1$

${{x}_{2}}=2$

${{y}_{1}}=5$

${{y}_{2}}=3$

$AB=\sqrt{{{\left( 1-2 \right)}^{2}}+{{\left( 5-3 \right)}^{2}}}$

$=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( 2 \right)}^{2}}}$

$=\sqrt{1+4}$

$=\sqrt{5}$

To find the distance between the points $B\left( 2,3 \right)$ and $C\left( -2,-11 \right)$

${{x}_{1}}=2$

${{x}_{2}}=-2$

${{y}_{1}}=3$

${{y}_{2}}=-11$

$BC=\sqrt{{{\left( 2-\left( -2 \right) \right)}^{2}}+{{\left( 3-\left( -11 \right) \right)}^{2}}}$

$=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 14 \right)}^{2}}}$

$=\sqrt{16+196}$

$=\sqrt{212}$

To find the distance between the points  $A\left( 1,5 \right)$  and  $C\left( -2,-11 \right)$

${{x}_{1}}=1$

${{x}_{2}}=-2$

${{y}_{1}}=5$

${{y}_{2}}=-11$

$CA=\sqrt{{{\left( 1-\left( -2 \right) \right)}^{2}}+{{\left( 5-\left( -11 \right) \right)}^{2}}}$

$=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 16 \right)}^{2}}}$

$=\sqrt{9+256}$

$=\sqrt{265}$

Since $AB+AC\ne BC$ and $AB\ne BC+AC$ 

and $AC\ne BC$

$\therefore $The points $A\left( 1,5 \right),B\left( 2,3 \right),C\left( -2,-11 \right)$ are not collinear.

4. Check whether $\left( 5,-2 \right),\left( 6,4 \right)$ and $\left( 7,-2 \right)$ are the vertices of an isosceles triangle.

Ans:

Given that,

Let the three points be $\left( 5,-2 \right),\left( 6,4 \right)$ and $\left( 7,-2 \right)$ are the vertices of the triangle.

To determine if the given points are the vertices of an isosceles triangle.

Let $A\left( 5,-2 \right),B\left( 6,4 \right),C\left( 7,-2 \right)$ be the vertices of the given triangle.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $A\left( 5,-2 \right)$ and $B\left( 6,4 \right)$

${{x}_{1}}=5$

${{x}_{2}}=6$

${{y}_{1}}=-2$

${{y}_{2}}=4$

$AB=\sqrt{{{\left( 5-6 \right)}^{2}}+{{\left( -2-4 \right)}^{2}}}$

$=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -6 \right)}^{2}}}$

$=\sqrt{1+36}$

$=\sqrt{37}$

To find the distance between the points $B\left( 6,4 \right)$ and $C\left( 7,-2 \right)$

${{x}_{1}}=6$

${{x}_{2}}=7$

${{y}_{1}}=4$

${{y}_{2}}=-2$

$BC=\sqrt{{{\left( 6-7 \right)}^{2}}+{{\left( 4-\left( -2 \right) \right)}^{2}}}$

$=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( 6 \right)}^{2}}}$

$=\sqrt{1+36}$

$=\sqrt{37}$

To find the distance between the points  $A\left( 5,-2 \right)$  and  $C\left( 7,-2 \right)$

${{x}_{1}}=5$

${{x}_{2}}=7$

${{y}_{1}}=-2$

${{y}_{2}}=-2$

$CA=\sqrt{{{\left( 5-7 \right)}^{2}}+{{\left( -2-\left( -2 \right) \right)}^{2}}}$

$=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 0 \right)}^{2}}}$

$=\sqrt{4+0}$

$=2$

We can conclude that $AB=BC$. 

Since two sides of the triangle are equal in length, $ABC$ is an isosceles triangle.

5. In a classroom, $4$ friends are seated at the points $A,B,C$ and $D$ are shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think $ABCD$ is a square?” Chameli disagrees.

Using the distance formula, find which of them is correct.

Ans:
Given that,

$4$ friends are seated at the points $A,B,C,D$

To find,

If they form square together by using distance formula

From the figure, we observe the points $A\left( 3,4 \right),B\left( 6,7 \right),C\left( 9,4 \right)$ and $D\left( 6,1 \right)$ are the positions of the four students.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $A\left( 3,4 \right)$ and $B\left( 6,7 \right)$

${{x}_{1}}=3$

${{x}_{2}}=6$

${{y}_{1}}=4$

${{y}_{2}}=7$

$AB=\sqrt{{{\left( 3-6 \right)}^{2}}+{{\left( 4-7 \right)}^{2}}}$

$=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -3 \right)}^{2}}}$

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$

To find the distance between the points $B\left( 6,7 \right)$ and $C\left( 9,4 \right)$

${{x}_{1}}=6$

${{x}_{2}}=9$

${{y}_{1}}=7$

${{y}_{2}}=4$

$BC=\sqrt{{{\left( 6-9 \right)}^{2}}+{{\left( 7-4 \right)}^{2}}}$

$=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 3 \right)}^{2}}}$

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$

To find the distance between the points  $C\left( 9,4 \right)$ and $\left( 6,1 \right)$

${{x}_{1}}=9$

${{x}_{2}}=6$

${{y}_{1}}=4$

$=2\left( lb+bh+lh \right)$

\[\]

$=2\left( 80 \right)CD=\sqrt{{{\left( 9-6 \right)}^{2}}+{{\left( 4-1 \right)}^{2}}}$

$\therefore =\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$

To find the distance between the points $A\left( 3,4 \right)$ and $D\left( 6,1 \right)$

${{x}_{1}}=3$

${{x}_{2}}=6$

${{y}_{1}}=4$

${{y}_{2}}=1$

$AB=\sqrt{{{\left( 3-6 \right)}^{2}}+{{\left( 4-1 \right)}^{2}}}$

$=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 3 \right)}^{2}}}$

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$

Since all sides of the squares are equal, now find the distance between the diagonals $AC$ and $BD$.

To find the distance between the points $A\left( 3,4 \right)$ and $C\left( 9,4 \right)$

$=13\text{ cm}{{x}_{1}}=3$

${{x}_{2}}=9$

${{y}_{1}}=4$

${{y}_{2}}=4$

Diagonal $AC=\sqrt{{{\left( 3-9 \right)}^{2}}+{{\left( 4-4 \right)}^{2}}}$

$=\sqrt{{{\left( -6 \right)}^{2}}+{{\left( 0 \right)}^{2}}}$

$=\sqrt{36+0}$

$=6$

Diagonal To find the distance between the points $B\left( 6,7 \right)$ and $D\left( 6,1 \right)$

${{x}_{1}}=6$

${{x}_{2}}=6$

${{y}_{1}}=7$

${{y}_{2}}=1$

Diagonal $BD=\sqrt{{{\left( 6-6 \right)}^{2}}+{{\left( 7-1 \right)}^{2}}}$

$=\sqrt{{{\left( 0 \right)}^{2}}+{{\left( 6 \right)}^{2}}}$

$=\sqrt{0+36}$

$=6$

Thus, the four sides \[\] and $DA$ are equal and its diagonals $AC$ and $BD$ are also equal.

$\therefore ABCD$ form a square and hence Champa was correct.

6. Name the type of quadrilateral forms, if any, by the following points, and give reasons for your answer.

(i) $\left( -1,-2 \right),\left( 1,0 \right),\left( -1,2 \right),\left( -3,0 \right)$-

Ans:

Given that,

Let the given points denote the vertices $A\left( -1,-2 \right),B\left( 1,0 \right),C\left( -1,2 \right),D\left( -3,0 \right)$ denote the vertices of the quadrilateral.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $A\left( -1,-2 \right)$ and $B\left( 1,0 \right)$

${{x}_{1}}=-1$

${{x}_{2}}=1$

${{y}_{1}}=-2$

${{y}_{2}}=0$

$AB=\sqrt{{{\left( -1-1 \right)}^{2}}+{{\left( -2-0 \right)}^{2}}}$

$=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -2 \right)}^{2}}}$

$=\sqrt{4+4}$

$=3.5\text{ cm}=\sqrt{8}$

$=2\sqrt{2}$

To find the distance between the points $B\left( 1,0 \right)$ and $C\left( -1,2 \right)$

${{x}_{1}}=1$

${{x}_{2}}=-1$

${{y}_{1}}=0$

${{y}_{2}}=2$

$BC=\sqrt{{{\left( 1-\left( -1 \right) \right)}^{2}}+{{\left( 0-2 \right)}^{2}}}$

$=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -2 \right)}^{2}}}$

$=\sqrt{4+4}$

$=\sqrt{8}$

$=2\sqrt{2}$

To find the distance between the points  $C\left( -1,2 \right)$ and \[\]

$=137.5{{x}_{1}}=-1$

${{x}_{2}}=-3$

${{y}_{1}}=2$

${{y}_{2}}=0$

$CD=\sqrt{{{\left( -1-\left( -3 \right) \right)}^{2}}+{{\left( 2-0 \right)}^{2}}}$

$=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 2 \right)}^{2}}}$

$=\sqrt{4+4}$

$=\sqrt{8}$

$=2\sqrt{2}$

To find the distance between the points $D\left( -3,0 \right)$ and $A\left( -1,-2 \right)$ 

${{x}_{1}}=-3$

${{x}_{2}}=-1$

${{y}_{1}}=0$

${{y}_{2}}=-2$

$AD=\sqrt{{{\left( -3-\left( -1 \right) \right)}^{2}}+{{\left( 0-\left( -2 \right) \right)}^{2}}}$

$=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 2 \right)}^{2}}}$

$=\sqrt{4+4}$

$=\sqrt{8}$

$=2\sqrt{2}$

To find the diagonals of the given quadrilateral

To find the distance between the points $A\left( -1,-2 \right)$ and $C\left( -1,2 \right)$

${{x}_{1}}=-1$

${{x}_{2}}=-1$

$=l{{y}_{1}}=-2$

${{y}_{2}}=2$

Diagonal $AC=\sqrt{{{\left( -1-\left( -1 \right) \right)}^{2}}+{{\left( -2-2 \right)}^{2}}}$

$=\sqrt{{{\left( 0 \right)}^{2}}+{{\left( -4 \right)}^{2}}}$

$=\sqrt{0+16}$

$=4$

To find the distance between the points $B\left( 1,0 \right)$ and $D\left( -3,0 \right)$

${{x}_{1}}=1$

${{x}_{2}}=-3$

${{y}_{1}}=0$

${{y}_{2}}=0$

Diagonal $BD=\sqrt{{{\left( 1-\left( -3 \right) \right)}^{2}}+{{\left( 0-0 \right)}^{2}}}$

$=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 0 \right)}^{2}}}$

$=\sqrt{16+0}$

$=4$

Since all sides of the given quadrilateral are of the same measure and the diagonals are also the same length. 

$\therefore $The given points of the quadrilateral form a square.

(ii) $\left( -3,5 \right),\left( 3,1 \right),\left( 0,3 \right),\left( -1,-4 \right)$-

Ans:

Given that,

Let the given points denote the vertices $A\left( -3,5 \right),B\left( 3,1 \right),C\left( 0,3 \right),D\left( -1,-4 \right)$ denote the vertices of the quadrilateral.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $A\left( -3,5 \right)$ and $B\left( 3,1 \right)$

${{x}_{1}}=-3$

${{x}_{2}}=3$

${{y}_{1}}=5$

${{y}_{2}}=1$

$AB=\sqrt{{{\left( -3-3 \right)}^{2}}+{{\left( 5-1 \right)}^{2}}}$

$=\sqrt{{{\left( -6 \right)}^{2}}+{{\left( 4 \right)}^{2}}}$

$=\sqrt{36+16}$

$=\sqrt{52}$

$=2\sqrt{13}$

To find the distance between the points \[\] and $C\left( 0,3 \right)$

${{x}_{1}}=3$

${{x}_{2}}=0$

${{y}_{1}}=1$

${{y}_{2}}=3$

$BC=\sqrt{{{\left( 3-0 \right)}^{2}}+{{\left( 1-3 \right)}^{2}}}$

$=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( -2 \right)}^{2}}}$

$=\sqrt{9+4}$

$=\sqrt{13}=14-5$

To find the distance between the points  $C\left( 0,3 \right)=45\pi $ and $\therefore D\left( -1,-4 \right)$

${{x}_{1}}=0$

${{x}_{2}}=-1$

${{y}_{1}}=3$

${{y}_{2}}=-4$

$CD=\sqrt{{{\left( 0-\left( -1 \right) \right)}^{2}}+{{\left( 3-\left( -4 \right) \right)}^{2}}}$

$=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 7 \right)}^{2}}}$

$=\sqrt{1+49}$

$=\sqrt{50}$

$=5\sqrt{2}$

To find the distance between the points $A\left( -3,5 \right)$ and $D\left( -1,-4 \right)$ 

${{x}_{1}}=-344\text{ }{{\text{m}}^{2}}$

${{x}_{2}}=-1$

${{y}_{1}}=5$

${{y}_{2}}=-4$

$AD=\sqrt{{{\left( -3-\left( -1 \right) \right)}^{2}}+{{\left( 5-\left( -4 \right) \right)}^{2}}}$

$=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 9 \right)}^{2}}}$

$=\sqrt{4+81}$

$=\sqrt{85}$

From the distance we found that,

$AB\ne BC\ne CD\ne AD$

By plotting the graph, we get,

From the graph above, we can note that the points $ABC$ are collinear.  

$\therefore $The quadrilateral cannot be formed by using the above points.

(iii) $\left( 4,5 \right),\left( 7,6 \right),\left( 4,3 \right),\left( 1,2 \right)$-

Ans:

Given that,

Let the given points denote the vertices $A\left( 4,5 \right),B\left( 7,6 \right),C\left( 4,3 \right),D\left( 1,2 \right)$ denote the vertices of the quadrilateral.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $A\left( 4,5 \right)$ and $B\left( 7,6 \right)$

${{x}_{1}}=4$

${{x}_{2}}=7$

${{y}_{1}}=5$

${{y}_{2}}=6$

$AB=\sqrt{{{\left( 4-7 \right)}^{2}}+{{\left( 5-6 \right)}^{2}}}$

$=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -1 \right)}^{2}}}$

$=\sqrt{9+1}$

$=\sqrt{10}$

To find the distance between the points $B\left( 7,6 \right)$ and $C\left( 4,3 \right)$

${{x}_{1}}=7$

${{x}_{2}}=4$

${{y}_{1}}=6$

${{y}_{2}}=3$

$BC=\sqrt{{{\left( 7-4 \right)}^{2}}+{{\left( 6-3 \right)}^{2}}}$

$=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 3 \right)}^{2}}}$

$=\frac{1.4}{2}=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$

To find the distance between the points $C\left(4,3 \right)$ and $D\left( 1,2 \right)$

${{x}_{1}}=4$

${{x}_{2}}=1$

${{y}_{1}}=3$

${{y}_{2}}=2$

$CD=\sqrt{{{\left( 4-1 \right)}^{2}}+{{\left( 3-2 \right)}^{2}}}$

$=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 1 \right)}^{2}}}$

$=\sqrt{9+1}$

$=\sqrt{10}$

To find the distance between the points x$A\left( 4,5 \right)$ and $D\left( 1,2 \right)$ 

${{x}_{1}}=4$

${{x}_{2}}=1$

${{y}_{1}}=5$

${{y}_{2}}=2$

$AD=\sqrt{{{\left( 4-1 \right)}^{2}}+{{\left( 5-2 \right)}^{2}}}$

$=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 3 \right)}^{2}}}$

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$

To find the diagonals of the given quadrilateral

To find the distance between the points $A\left( 4,5 \right)$ and $C\left( 4,3 \right)$

${{x}_{1}}=4$

${{x}_{2}}=4$

${{y}_{1}}=5$

${{y}_{2}}=3$

Diagonal $AC=\sqrt{{{\left( 4-4 \right)}^{2}}+{{\left( 5-3 \right)}^{2}}}$

$=\sqrt{{{\left( 0 \right)}^{2}}+{{\left( 2 \right)}^{2}}}$

$=\sqrt{0+4}$

$=2$

To find the distance between the points $B\left( 7,6 \right)$ and $D\left( 1,2 \right)$

${{x}_{1}}=7$

${{x}_{2}}=1$

${{y}_{1}}=6$

${{y}_{2}}=2$

Diagonal $BD=\sqrt{{{\left( 7-1 \right)}^{2}}+{{\left( 6-2 \right)}^{2}}}$

$=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( 4 \right)}^{2}}}$

$=\sqrt{36+16}$

$=\sqrt{52}$

$=2\sqrt{13}$

From the above calculation, the opposite sides of the quadrilateral are of same length and the diagonals are not of same length

$\therefore $The given points of the quadrilateral form a parallelogram.

Question 7:

Find the point on the $x$-axis which is equidistant from $\left( 2,-5 \right)$ and $\left( -2,9 \right)$.

Ans:

Given that,

$\left( 2,-5 \right)$

$\left( -2,9 \right)$

To find,

The point that is equidistant from the points $\left( 2,-5 \right)$ and $\left( -2,9 \right)$

Let us consider the points as $A\left( 2,-5 \right)$ and $B\left( -2,9 \right)$ and to find the equidistant point $P$.

Since the point is on $x$-axis, the coordinates of the required point is of the form $P\left( x,0 \right)$.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

To find the distance between the points $P\left( x,0 \right)$ and $A\left( 2,-5 \right)$ 

${{x}_{1}}=x$

${{x}_{2}}=2$

${{y}_{1}}=0$

${{y}_{2}}=-5$

$PA=\sqrt{{{\left( x-2 \right)}^{2}}+{{\left( 0-\left( -5 \right) \right)}^{2}}}$

$=\sqrt{{{\left( x-2 \right)}^{2}}+25}$

To find the distance between the points $A\left( x,0 \right)$ and $B\left( -2,9 \right)$

${{x}_{1}}=x$

${{x}_{2}}=-2$

${{y}_{1}}=0$

${{y}_{2}}=9$

$PB=\sqrt{{{\left( x-\left( -2 \right) \right)}^{2}}+{{\left( 0-9 \right)}^{2}}}$

$=\sqrt{{{\left( x+2 \right)}^{2}}+{{\left( -9 \right)}^{2}}}$

$=\sqrt{{{\left( x+2 \right)}^{2}}+81}$

Since the distance are equal in measure,

$PA=PB$

$\sqrt{{{\left( x-2 \right)}^{2}}+25}$ $=\sqrt{{{\left( x+2 \right)}^{2}}+81}$

Taking square on both sides,

${{(x-2)}^{2}}+25= ${{(x+2)}^{2}}+81

By solving, we get,

${{x}^{2}}+4-4x+25={{x}^{2}}+4+4x+81$

$8x=-56$

$x=-7$

The coordinate is $\left( -7,0 \right)$.

$\therefore $The point that is equidistant from $\left( 2,-5 \right)$ and $\left( -2,9 \right)$ is $\left( -7,0 \right)$.

8. Find the values of $y$for which the distance between the points $P\left( 2,-3 \right)$ and $Q\left( 10,y \right)$ is $10$units.

Ans:

Given that,

$P\left( 2,-3 \right)$

$Q\left( 10,y \right)$

The distance between these points is $10$ units.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

.To find the distance between the points $P\left( 2,-3 \right)$ and $Q\left( 10,y \right)$

${{x}_{1}}=2$

${{x}_{2}}=10$

${{y}_{1}}=-3$

${{y}_{2}}=y$

$PQ=\sqrt{{{\left( 10-2 \right)}^{2}}+{{\left( y-\left( -3 \right) \right)}^{2}}}$

$=\sqrt{{{\left( -8 \right)}^{2}}+{{\left( y+3 \right)}^{2}}}$

$=\sqrt{64+{{\left( y+3 \right)}^{2}}}$

Since the distance between them is $10$ units,

$\sqrt{64+{{\left( y+3 \right)}^{2}}}=10$

Squaring on both sides, we get,

$64+{{\left( y+3 \right)}^{2}}=100$

${{\left( y+3 \right)}^{2}}=36$

$y+3=\pm 6$

So, $y+3=6$

$y=3$

And, $y+3=-6$

$y=-9$

$\therefore $The possible values of $y$ are $y=3$ or $y=-9$.

9. If $Q\left( 0,1 \right)$ is equidistant from $P\left( 5,-3 \right)$ and $R\left( x,6 \right)$, find the values of $x$. Also find the distance of $QR$ and $PR$.

Ans:

Given that,

$Q\left( 0,1 \right)$

$P\left( 5,-3 \right)$

$R\left( x,6 \right)$

To find,

• The values of $x$

• The distance of $QR$ and $PR$

$Q$ is equidistant between $P$ and $R$.

SO $PQ=QR$

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

$\sqrt{{{\left( 5-0 \right)}^{2}}+{{\left( -3-1 \right)}^{2}}}=\sqrt{{{\left( 0-x \right)}^{2}}+{{\left( 1-6 \right)}^{2}}}$

$\sqrt{{{\left( 5 \right)}^{2}}+{{\left( -4 \right)}^{2}}}=\sqrt{{{\left( -x \right)}^{2}}+{{\left( -5 \right)}^{2}}}$

$\sqrt{25+16}=\sqrt{{{x}^{2}}+25}$

Squaring on both sides, we get,

$41={{x}^{2}}+25$

${{x}^{2}}=16\pi $

 x = ±4

Thus the point $R$ is $R\left( 4,6 \right)$ or $\left( -4,6 \right)$.

To find the distance $PR$ and $QR$

Case (1):

When the point is $R\left( 4,6 \right)$

Distance between the point $P\left( 5,-3 \right)$ and $R\left( 4,6 \right)$ is,

$PR=\sqrt{{{\left( 5-4 \right)}^{2}}+{{\left( -3-6 \right)}^{2}}}$

$=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( -9 \right)}^{2}}}$

$=\sqrt{1+81}$

$=\sqrt{82}$

Distance between the point $Q\left( 0,1 \right)$ and $R\left( 4,6 \right)$,

$QR=\sqrt{{{\left( 0-4 \right)}^{2}}+{{\left( 1-6 \right)}^{2}}}$

$=\sqrt{{{\left( -4 \right)}^{2}}+{{\left( -5 \right)}^{2}}}$

$=\sqrt{16+25}$

$=\sqrt{41}$

Case (2):

When the point is $R\left( -4,6 \right)$

Distance between the point $P\left( 5,-3 \right)$ and $R\left( 4,6 \right)$,

$PR=\sqrt{{{\left( 5-\left( -4 \right) \right)}^{2}}+{{\left( -3-6 \right)}^{2}}}$

$=\sqrt{{{\left( 9 \right)}^{2}}+{{\left( -9 \right)}^{2}}}$

$=\sqrt{81+81}$

$=\sqrt{162}$

$=9\sqrt{2}$

Distance between the point $Q\left( 0,1 \right)$ and $R\left( 4,6 \right)$,

$QR=\sqrt{{{\left( 0-\left( -4 \right) \right)}^{2}}+{{\left( 1-6 \right)}^{2}}}$

$=\sqrt{{{\left( 4 \right)}^{3}}+{{\left( -5 \right)}^{2}}}$

$=\sqrt{16+25}$

$=\sqrt{41}$

10. Find a relation between $x$ and $y$ such that the point $\left( x,y \right)$ is equidistant from the point $\left( 3,6 \right)$ and $\left( -3,4 \right)$.

Ans:

Given that,

$\left( x,y \right)$ is equidistant from $\left( 3,6 \right)$ and $\left( -3,4 \right)$

To find,

The value of $\left( x,y \right)$

Let $P\left( x,y \right)$ is equidistant from $A\left( 3,6 \right)$ and $B\left( -3,4 \right)$

Since they are equidistant,

$PA=PB$

$=66\text{ c}{{\text{m}}^{3}}\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y-6 \right)}^{2}}}=\sqrt{\left( x-{{\left( -3 \right)}^{2}} \right)+{{\left( y-4 \right)}^{2}}}$

$\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y-6 \right)}^{2}}}=\sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}}$

Squaring on both sides, we get,

${{\left( x-3 \right)}^{2}}+{{\left( y-6 \right)}^{2}}={{\left( x+3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}$

${{x}^{2}}+9-6x+{{y}^{2}}+36-12y={{x}^{2}}+9+6x+{{y}^{2}}+16-8y$

$36-16=12x+4y$

$3x+y=5$

$3x+y-5=0$

$\therefore $The relation between $x$ and $y$ is given by $3x+y-5=0$

Exercise (7.2)

1.Find the coordinates of the point which divides the join of $\left( -1,7 \right)$ and $\left( 4,-3 \right)$ in the ratio $2:3$

Ans:

Given that,

The points $A\left( -1,7 \right)$ and $B\left( 4,-3 \right)$

Ratio $m:n=$ $2:3$

To find,

The coordinates

Let $P\left( x,y \right)$ be the required coordinate

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$P\left( x,y \right)=\left[ \frac{2\left( 4 \right)+3\left( -1 \right)}{2+3},\frac{2\left( -3 \right)+3\left( 7 \right)}{2+3} \right]$

$=\left[ \frac{8-3}{5},\frac{-6+21}{5} \right]$

$=\left[ \frac{5}{5},\frac{15}{5} \right]$

$=\left( 1,3 \right)$

$\therefore $The coordinates of $P$ is $P\left( 1,3 \right)$.

2. Find the coordinates of the point of trisection of the line segment joining $\left( 4,-1 \right)$ and $\left( -2,-3 \right)$.

Ans:

Given that,

The line segment joining $\left( 4,-1 \right)$ and $\left( -2,-3 \right)$

To find,

The coordinates of the point 

Let the line segment joining the points be $A\left( 4,-1 \right)$ and $B\left( -2,-3 \right)$

Let $P\left( {{x}_{1}},{{y}_{1}} \right)$ and $Q\left( {{x}_{2}},{{y}_{2}} \right)$ are the points of trisection of the line segment joining the points 

$AP=PQ=QB$

From the diagram, the point $P$ divides $AB$ internally in the ratio of $1:2$

Hence $m:n=1:2$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$P\left( {{x}_{1}},{{y}_{1}} \right)=\left[ \frac{1\left( -2 \right)+1\left( 4 \right)}{1+2},\frac{1\left( -3 \right)+2\left( -1 \right)}{1+2} \right]$

$=\left[ \frac{-2+8}{3},\frac{-3-2}{3} \right]$

$=\left[ \frac{6}{3},\frac{-5}{3} \right]$

$=\left( 2,-\frac{5}{3} \right)$

$\therefore $The coordinates of $P$ is $P\left( 2,-\frac{5}{3} \right)$.

From the diagram, the point $Q$ divides $AB$ internally in the ratio of $2:1$

Hence $m:n=2:1$

By section formula,

$Q\left( {{x}_{2}},{{y}_{2}} \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$Q\left( {{x}_{2}},{{y}_{2}} \right)=\left[ \frac{2\left( -2 \right)+1\left( 4 \right)}{2+1},\frac{2\left( -3 \right)+3\left( -1 \right)}{2+1} \right]$

$=\left[ \frac{-4+4}{3},\frac{-6-1}{3} \right]$

$=\left[ 0,-\frac{7}{3} \right]$

$\therefore $The coordinates of $Q$ is $Q\left( 0,-\frac{7}{3} \right)$.

3. To conduct Sports day activities, in your rectangular shaped school ground $ABCD$, lines have been drawn with chalk powder at a distance of $1\text{ m}$ each. $100$ flower pots have been placed at a distance of $1\text{ m}$ from each other along $AD$, as shown in the following figure. Niharika runs $\frac{1}{4}\text{ th}$ the distance $AD$ on the $2\text{nd}$ line and posts a green flag. Preet runs $\frac{1}{5}\text{ th}$the distance $AD$ on the eighth line and posts a red flag. What is the distance between both the flags?If Rashmi has to post a blue flag exactly halfway between the line segments joining the two flags, where should she post her flag?

Ans:

Given that,

• Niharika posted her green flag at a distance of $P$which is $\frac{1}{4}\times 100=25\text{ m}$from the starting point of the second line. The coordinate of $P$ is $P\left( 2,25 \right)$

• Preet posted red flag at $\frac{1}{5}$ of a distance $Q$ which is $\frac{1}{5}\times 100=20\text{ m}$from the starting point of the eighth line. The coordinate of $Q$ is $\left( 8,20 \right)$

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

The distance between the flags $P\left( 2,25 \right),Q\left( 8,20 \right)$ is given by,

$PQ=\sqrt{{{\left( 8-2 \right)}^{2}}+{{\left( 25-20 \right)}^{2}}}$

$=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( 5 \right)}^{2}}}$

$=\sqrt{36+25}$

$=\sqrt{61}\text{ m}$

Rashmi should post her blue flag in the mid-point of the line joining points $P$ and $Q$. Let this point be $M\left( x,y \right)$.

By section formula,

$M\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$M\left( x,y \right)=\left[ \frac{1\left( 2 \right)+1\left( 8 \right)}{1+1},\frac{1\left( 25 \right)+1\left( 20 \right)}{1+1} \right]$

$=\left[ \frac{2+8}{2},\frac{25+20}{2} \right]$

$=\left[ \frac{10}{2},\frac{45}{2} \right]$

$=\left( 5,22.5 \right)$

$\therefore $Rashmi should place her blue flag at $22.5\text{ m}$ in the fifth line.

4. Find the ratio in which the line segment joining the points $\left( -3,10 \right)$ and $\left( 6,-8 \right)$ is divided by $\left( -1,6 \right)$

Ans:

Given that,

The line segment $\left( -3,10 \right)$ and $\left( 6,-8 \right)$

The point $\left( -1,6 \right)$ divides the line segment

To find,

The ratio of dividing line segment 

Let the line segment be $A\left( -3,10 \right)$ and $B\left( 6,-8 \right)$ divided by point $P\left( -1,6 \right)$in the ratio of $k:1$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

Equating the $x$term,

$-1=\frac{6k-3}{k+1}$

$-k-1=6k-3$

$7k=2$

$k=\frac{2}{7}$

$\therefore $The point $P$ divides the line segment $AB$ in the ratio of $2:7$

5. Find the ratio in which the line segment joining $A\left( 1,-5 \right)$ and $B\left( -4,5 \right)$ is divided by the $x$ axis. Also find the coordinates of the point of division.

Ans:

Given that,

The line segment joining the points $A\left( 1,-5 \right)$ and $B\left( -4,5 \right)$

To find,

• The ratio

• The coordinates of the point of division

Let the ratio be $k:1$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$P\left( x,y \right)=\left[ \frac{k\left( -4 \right)+1\left( 1 \right)}{k+1},\frac{k\left( 5 \right)+1\left( -5 \right)}{k+1} \right]$

$=\left[ \frac{-4k+1}{k+1},\frac{5k-5}{k+1} \right]$

We know that $y$ coordinate on $x$ axis is zero.

$\frac{5k-5}{k+1}=0$

$5k-5=0$

$5k=5$

$k=1$

Therefore $x$ axis divides it in the ratio of $1:1$

Division point, $P=\left( \frac{-4\left( 1 \right)+1}{1+1},\frac{5\left( 1 \right)-5}{1+1} \right)$

$=\left( \frac{-4+1}{2},\frac{5-5}{2} \right)$

$=\left( -\frac{3}{2},0 \right)$

$\therefore $The ratio at which the line segment is divided is $1:1$ and the point of division is  $\left( -\frac{3}{2},0 \right)$.

6. If $\left( 1,2 \right),\left( 4,y \right),\left( x,6 \right)$ and $\left( 3,5 \right)$ are the vertices of the parallelogram taken in order, find $x$ and $y$.

Ans:

Given that,

The vertices of the parallelogram are $A\left( 1,2 \right),B\left( 4,y \right),C\left( x,6 \right),D\left( 3,5 \right)$

To find,

The value of $x$ and $y$

• The diagonals of the parallelogram bisect each other at $O$. 

• Intersection point $O$ of diagonal $AC$ and $BD$divides these diagonals. So $O$ is the midpoint of $AC$ and $BD$.

If $O$ is the midpoint of $AC$,

$O=\left( \frac{1+x}{2},\frac{2+6}{2} \right)$

$=\left( \frac{1+x}{2},\frac{8}{2} \right)$

$=\left( \frac{1+x}{2},4 \right)$

If $O$ is the midpoint of $BD$,

$O=\left( \frac{4+3}{2},\frac{5+y}{2} \right)$

$=\left( \frac{7}{2},\frac{5+y}{2} \right)$

Equating the points of $O$,

$\frac{x+1}{2}=\frac{7}{2}$ and 

$4=\frac{5+y}{2}$

Finding $x$ term,

$\frac{x+1}{2}=\frac{7}{2}$

$x+1=7$

$x=6$

Finding $y$ term,

$4=\frac{5+y}{2}$

$5+y=8$

$y=3$

The value of $x$ and $y$ are $x=6$ and $y=3$

7. Find the coordinates of point $A$, where $AB$ is the diameter of circle whose center is $\left( 2,-3 \right)$ and $B$ is $\left( 1,4 \right)$

Ans:

Given that,

• Center is $C\left( 2,-3 \right)$

• The coordinate of $B$ is $B\left( 1,4 \right)$

To find,

The coordinate of $A$

Let the coordinate of $A$ be $A\left( x,y \right)$

Midpoint of $AB$ is $C\left( 2,-3 \right)$and so,

$\left( 2,-3 \right)=\left( \frac{x+1}{2},\frac{y+4}{2} \right)$

Equating $x$ term,

$\frac{x+1}{2}=2$

$x+1=4$

$x=3$

Equating $y$ term,

$\frac{y+4}{2}=-3$

$y+4=-6$

$y=-10$

$\therefore $The coordinate of $A$ is $A\left( 3,-10 \right)$

8. If $A$ and $B$ are $\left( -2,-2 \right)$ and $\left( 2,-4 \right)$ respectively, find the coordinates of $P$ such that $AP=\frac{3}{7}AB$ and $P$ lies on the line segment $AB$.

Ans:

Given that,

• The coordinates are $A\left( -2,-2 \right)$ and $B\left( 2,-4 \right)$

• $AP=\frac{3}{7}AB$

To find, 

The coordinate of $P$

$AP=\frac{3}{7}AB$

$\frac{AB}{AP}=\frac{7}{3}$

From the figure, $AB=AP+PB$

$\frac{AP+PB}{AP}=\frac{3+4}{3}$

$1+\frac{PB}{AP}=1+\frac{4}{3}$

$\frac{PB}{AP}=\frac{4}{3}$

$\therefore AP:PB=3:4$

Thus, the point $P\left( x,y \right)$ divides the line segment $AB$ in the ratio of $3:4$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$P\left( x,y \right)=\left[ \frac{3\left( 2 \right)+4\left( -2 \right)}{3+4},\frac{3\left( -4 \right)+4\left( -2 \right)}{3+4} \right]$

$=\left[ \frac{6-8}{7},\frac{-12-8}{7} \right]$

$=\left[ -\frac{2}{7},-\frac{20}{7} \right]$

$\therefore $The coordinates of $P$ is $P\left( -\frac{2}{7},-\frac{20}{7} \right)$.

7. Find the coordinates of the points which divide the line segment joining $A\left( -2,2 \right)$ and $B\left( 2,8 \right)$ into four equal parts.

Ans;

Given that,

The line segment $A\left( -2,2 \right)$ and $B\left( 2,8 \right)$

To find,

The coordinate that divides the line segment into four equal parts

Form the figure, ${{P}_{1}},{{P}_{2}},{{P}_{3}}$ be the points that divide the line segment $AB$ into four equal parts.

Point ${{P}_{1}}$ divides the line segment $AB$ in the ratio of $1:3$, so, 

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

${{P}_{1}}\left( x,y \right)=\left[ \frac{1\left( 2 \right)+3\left( -2 \right)}{1+3},\frac{1\left( 8 \right)+3\left( 2 \right)}{1+3} \right]$

$=\left[ \frac{-4}{4},\frac{14}{4} \right]$

$=\left( -1,\frac{7}{2} \right)$

Point ${{P}_{2}}$ divides the line segment $AB$ in the ratio of $1:1$, so, 

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

${{P}_{2}}\left( x,y \right)=\left[ \frac{1\left( 2 \right)+1\left( -2 \right)}{1+1},\frac{1\left( 8 \right)+1\left( 2 \right)}{1+1} \right]$

$=\left[ \frac{2+\left( -2 \right)}{2},\frac{2+8}{2} \right]$

$=\left( 0,5 \right)$

Point ${{P}_{3}}$ divides the line segment $AB$ in the ratio of $3:1$, so, 

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

${{P}_{3}}\left( x,y \right)=\left[ \frac{3\left( 2 \right)+1\left( -2 \right)}{3+1},\frac{3\left( 8 \right)+1\left( 2 \right)}{3+1} \right]$

$=\left[ \frac{6-2}{4},\frac{24+2}{4} \right]$

$=\left( 1,\frac{13}{2} \right)$

The coordinates that divide the line segment into four equal parts are ${{P}_{1}}\left( -1,\frac{7}{2} \right),{{P}_{2}}\left( 0,5 \right)$ and ${{P}_{3}}\left( 1,\frac{13}{2} \right)$

10. Find the area of the rhombus if its vertices are $\left( 3,0 \right),\left( 4,5 \right),\left( -1,4 \right)$ and $\left( -2,-1 \right)$ taken in order. (Hint: Area of a rhombus = $\frac{1}{2}$(Product of its diagonals))

Ans:

Given that,

The vertices of the rhombus are $A\left( 3,0 \right),B\left( 4,5 \right),C\left( -1,4 \right)$ and $D\left( -2,-1 \right)$

To find, 

The area of the rhombus

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

Distance between the diagonal $AC$ is given by,

$AC=\sqrt{{{\left( 3-\left( -1 \right) \right)}^{2}}+{{\left( 0-4 \right)}^{2}}}$

$=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( -4 \right)}^{2}}}$

$=\sqrt{16+16}$

$=\sqrt{32}$

$=4\sqrt{2}$

Distance between the diagonal $BD$ si given by,

$BD=\sqrt{{{\left( 4-\left( -2 \right) \right)}^{2}}+{{\left( 5-\left( -1 \right) \right)}^{2}}}$

$=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( 6 \right)}^{2}}}$

$=\sqrt{36+36}$

$=\sqrt{72}$

$=6\sqrt{2}$

Area of the rhombus = $\frac{1}{2}\times $(Products of lengths of diagonals

$=\frac{1}{2}\times AC\times BD$

$=\frac{1}{2}\times 4\sqrt{2}\times 6\sqrt{2}$

$=24$ square units

Exercise (7.3)

1.Find the area of the triangle whose vertices are 

(i) $\left( 2,3 \right),\left( -1,0 \right),\left( 2,-4 \right)$

Ans:

Given that,

The vertices of the triangle whose points are $A\left( 2,3 \right),B\left( -1,0 \right),C\left( 2,-4 \right)$

To find,

The area of the triangle

Area of the triangle $=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

Substitute the value of ${{x}_{1}},{{x}_{2}},{{x}_{3}}$ in the above formula,

Area of given triangle$=\frac{1}{2}\left[ 2\left( 0-\left( -4 \right) \right)+-1\left( -4-3 \right)+2\left( 3-0 \right) \right]$

$=\frac{1}{2}\left( 8+7+6 \right)$

$=\frac{21}{2}$ square units

Area of the triangle with vertices $A\left( 2,3 \right),B\left( -1,0 \right),C\left( 2,-4 \right)$ is $\frac{21}{2}$ square units

(ii) $\left( -5,-1 \right),\left( 3,-5 \right),\left( 5,2 \right)$

Ans:

Given that,

The vertices of the triangle whose points are $A\left( -5,-1 \right),B\left( 3,-5 \right),C\left( 5,2 \right)$

To find,

The area of the triangle

Area of the triangle $=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

Substitute the value of ${{x}_{1}},{{x}_{2}},{{x}_{3}}$ in the above formula,

Area of given triangle$=\frac{1}{2}\left[ -5\left( -5-\left( -2 \right) \right)+3\left( 2-\left( -1 \right) \right)+5\left( -1-\left( -5 \right) \right) \right]$

$=\frac{1}{2}\left( 35+9+20 \right)$

$=32$ square units

Area of the triangle with vertices $A\left( -5,-1 \right),B\left( 3,-5 \right),C\left( 5,2 \right)$ is $32$ square units.

.

2. In each of the following find the value of ‘k’, for which the points are collinear.

(i) $\left( 7,-2 \right),\left( 5,1 \right),\left( 3,-k \right)$

Ans:

Given that, 

The vertices of the triangle are $A\left( 7,-2 \right),B\left( 5,1 \right),C\left( 3,-k \right)$

The points are collinear

To find,

The value of $k$

If the points are collinear, then,

Area of the triangle $=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]=0$

Substitute the value of ${{x}_{1}},{{x}_{2}},{{x}_{3}}$ in the above equation,

$\frac{1}{2}\left[ 7\left( 1-k \right)+5\left( k-\left( -2 \right) \right)+3\left( -2-1 \right) \right]=0$

$7-7k+5k+10-9=0$

$-2k+8=0$

$2k=8$

$k=4$

$\therefore $The value of $k$ is $4$ 

(ii) $\left( 8,1 \right),\left( k,-4 \right),\left( 2,-5 \right)$

Ans:

Given that, 

The vertices of the triangle are $A\left( 8,1 \right),B\left( k,-4 \right),C\left( 2,-5 \right)$

The points are collinear

To find,

The value of $k$

If the points are collinear, then,

Area of the triangle $=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]=0$

Substitute the value of ${{x}_{1}},{{x}_{2}},{{x}_{3}}$ in the above equation,

$\frac{1}{2}\left[ 8\left( -4-\left( -5 \right) \right)+k\left( -5-1 \right)+2\left( 1-\left( -4 \right) \right) \right]=0$

$8-6k+10=0$

$6k=18$

$k=3$

$\therefore $The value of $k$ is $3$ 

3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are $\left( 0,-1 \right),\left( 2,1 \right)$ and $\left( 0,3 \right)$. Find the ratio of this area to the area of the given triangle.

Ans:

Given that,

The vertices of the triangle are $A\left( 0,-1 \right),B\left( 2,1 \right),C\left( 0,3 \right)$

To find,

The ratio of areas of two triangles

Area of the triangle $=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

Area of the triangle $ABC=\frac{1}{2}\left[ 0\left( 1-3 \right)+2\left( 3-\left( -1 \right) \right)+0\left( -1-1 \right) \right]$

$=\frac{1}{2}\left( 8 \right)$

$=4$square units

A triangle is formed by joining the midpoints of the triangle ABC.

Let $D,E,F$ be the midpoints of the sides $AB,BC,CA$ of the triangle respectively.

$D=\left( \frac{0+2}{2},\frac{-1+!}{2} \right)$

$=\left( 1,0 \right)$

$E=\left( \frac{0+0}{2},\frac{3-1}{2} \right)$

$=\left( 0,1 \right)$

$F=\left( \frac{2+0}{2},\frac{1+3}{2} \right)$

$=\left( 1,2 \right)$

Area of triangle DEF$=\frac{1}{2}\left[ 1\left( 2-1 \right)+1\left( 1-0 \right)+0\left( 0-2 \right) \right]$

$=\frac{1}{2}\left( 2 \right)$

$=1$ square unit

$\therefore $Ratio of the triangle $DEF$ to the ratio of the triangle $ABC$ is $1:4$

4. Find the area of the quadrilateral whose vertices, taken in order, are $\left( -4,-2 \right),\left( -3,-5 \right),\left( 3,-2 \right)$ and $\left( 2,3 \right)$

Ans:

Given that,

The vertices of the triangle are $A\left( -4,-2 \right),B\left( -3,-5 \right),C\left( 3,-2 \right),D\left( 2,3 \right)$

To find,

The area of the quadrilateral $ABCD$

Join $AC$ so that it forms two triangles $\vartriangle ABC$ and $\vartriangle ACD$

Area of the triangle $=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

Substituting the value of $A\left( -4,-2 \right),B\left( -3,-5 \right),C\left( 3,-2 \right)$ in the above formula,

Area of $\vartriangle ABC=\frac{1}{2}\left[ -4\left( -5-\left( -2 \right) \right)+-3\left( -2-\left( -2 \right) \right)+3\left( -2-\left( -2 \right) \right) \right]$

$=\frac{1}{2}\left[ 12+0+9 \right]$

$=\frac{21}{2}$ square unit

Substituting the value of $A\left( -4,-2 \right),C\left( 3,-2 \right),D\left( 2,3 \right)$ in the above formula,

Area of $\vartriangle ACD=\frac{1}{2}\left[ -4\left( -2-\left( -3 \right) \right)+3\left( 3-\left( -2 \right) \right)+2\left( -2-\left( -2 \right) \right) \right]$

$=\frac{1}{2}\left[ 20+15+0 \right]$

$=\frac{35}{2}$ square unit

Area of quadrilateral $ABCD=$Area of $\vartriangle ABC+$Area of $\vartriangle ACD$

$=\frac{21}{2}+\frac{35}{2}$

$=\frac{56}{2}$

$=28$ square unit

The area of the quadrilateral $ABCD$ is $28$ square units.

5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for $\vartriangle ABC$ whose vertices are $A\left( 4,-6 \right),B\left( 3,-2 \right)$ and $C\left( 5,2 \right)$

Ans:

Given that,

The vertices of the triangle are $A\left( 4,6 \right),B\left( 3,-2 \right),C\left( 5,2 \right)$

To verify,

The median of the triangle divides the triangle into two triangles of equal areas.

Let $D$ be the midpoint of $BC$.

$D=\left( \frac{3+5}{2},\frac{-2+2}{2} \right)$

$=\left( \frac{8}{2},0 \right)$

$=\left( 4,0 \right)$

$AD$ is the median of the $\vartriangle ABC$ and it divides the triangle into two triangles.

Area of the triangle $=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

Substituting the values of $A\left( 4,6 \right),B\left( 3,-2 \right)$ and $D\left( 4,0 \right)$ in the above formula,

Area of $\vartriangle ABD=\frac{1}{2}\left[ 4\left( -2-0 \right)+3\left( 0-\left( -6 \right) \right)+\left( -6-\left( -2 \right) \right) \right]$

$=\frac{1}{2}\left[ -8+18-16 \right]$

$=\frac{1}{2}\left( 6 \right)$

$=3$ square unit

Substituting the values of $A\left( 4,6 \right),D\left( 4,0 \right),C\left( 5,2 \right)$ in the above formula,

Area of $\vartriangle ADC=\frac{1}{2}\left[ 4\left( 0-\left( -2 \right) \right)+4\left( 2-\left( -6 \right) \right)+5\left( -6-0 \right) \right]$

$=\frac{1}{2}\left( -8+32-30 \right)$

$=-3$ square unit

Area cannot be in negative value. So ignore the negative value.

Both the areas are equal.

$\therefore $Median $AD$ of the triangle divides the triangle into two triangles of equal areas

Exercise (7.4)

1.Determine the ratio in which the line $2x+y-4=0$ divides the line segment joining the points $A\left( 2,-2 \right)$and $B\left( 3,7 \right)$.

Ans:

Given that,

• A line x$2x+y-4=0$

• A line segment joining the point $A\left( 2,-2 \right)$ and $B\left( 3,7 \right)$

To determine,

The ratio

The given line $2x+y-4=0$ divides the line segment joining the points $A\left( 2,-2 \right)$ and $B\left( 3,7 \right)$ in a ratio of $k:1$ at the point $C$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$C\left( x,y \right)=\left[ \frac{k\left( 3 \right)+1\left( 2 \right)}{k+1},\frac{k\left( 7 \right)+1\left( -2 \right)}{k+1} \right]$

$=\left[ \frac{3k+2}{k+1},\frac{7k-2}{k+1} \right]$

The point $C$ also lies on the line $2x+y-4=0$

Substituting the value of $C\left( x,y \right)$ in the given line, we get,

$2\left( \frac{3k+2}{k+1} \right)+\left( \frac{7k-2}{k+1} \right)-4=0$

By solving this, we get,

$\frac{6k+4+7k-2-4k-4}{k+1}=0$

$9k-2=0$

$k=\frac{2}{9}$

$\therefore $The ratio in which the line $2x+y-4=0$ divides the line joining two points $A\left( 2,-2 \right)$ and $B\left( 3,7 \right)$ is $2:9$ internally.

2. Find a relation between $x$ and $y$ if the points $\left( x,y \right),\left( 1,2 \right)$ and $\left( 7,0 \right)$ are collinear.

Ans:

Given that,

The points $A\left( x,y \right),B\left( 1,2 \right),C\left( 7,0 \right)$

The points are collinear

To find,

The relation between $x$ and $y$

Since the given points are collinear, then the area of the triangle formed by these points is equal to zero.

Area of the triangle $=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

Substituting the above points in the above formula, w get,

Area $=\frac{1}{2}\left[ x\left( 2-0 \right)+1\left( 0-y \right)+7\left( y-2 \right) \right]$

$\frac{1}{2}\left[ 2x-y+7y-14 \right]=0$

$2x+6y-14=0$

$x+3y-7=0$

$\therefore $The relation between $x$ and $y$ is $x+3y-7=0$

3. Find the center of the circle passing through the points $\left( 6,-6 \right),\left( 3,-7 \right)$ and $\left( 3,3 \right)$.

Ans:

Given that,

Center passing through the points $A\left( 6,-6 \right),B\left( 3,-7 \right),C\left( 3,3 \right)$

To find,

The center of the circle

Let $O\left( x,y \right)$ be the center of the circle

Distance from center $O$to the points $A,B,C$ is the same since they form the radius of the circle.

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

$OA=\sqrt{{{\left( x-6 \right)}^{2}}+{{\left( y+6 \right)}^{2}}}$

$OB=\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y+7 \right)}^{2}}}$

$OC=\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y-3 \right)}^{2}}}$

Equate $OA=OB$ since it represents radius

$\sqrt{{{\left( x-6 \right)}^{2}}+{{\left( y+6 \right)}^{2}}}=\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y+7 \right)}^{2}}}$

Squaring on both sides and solving them, we get,

${{x}^{2}}+36-12x+{{y}^{2}}+36+12y={{x}^{2}}+9-6x+{{y}^{2}}+49+14y$

$-6x-2y+14=0$

$3x+y=7$                                             ……(1)

Similarly, $OA=OC$ since it represents radius

$\sqrt{{{\left( x-6 \right)}^{2}}+{{\left( y+6 \right)}^{2}}}=\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y-3 \right)}^{2}}}$

Squaring on both sides and solving them, we get,

${{x}^{2}}+36-12x+{{y}^{2}}+36+12y={{x}^{2}}+9-6x+{{y}^{2}}+9-6y$

$-6x+18y+54=0$

$-3x+9y=-27$                                      ……(2)

By adding (1) and (2), w get,

$10y=-20$

$y=-2$

Substituting $y=-2$ in the equation (1), we get,

$3x-2=7$

$3x=9$

$x=3$

$\therefore $The center of the circle is $C\left( 3,-2 \right)$

.

4. The two opposite vertices of a square are $\left( -1,2 \right)$ and $\left( 3,2 \right)$. Find the coordinates of the two other two vertices

Ans:

Given that,

The two opposite vertices of the square are $A\left( 1,-2 \right)$ and $C\left( 3,2 \right)$

To find,

The other two coordinates

Let $ABCD$ be a square having the vertices $A\left( -1,2 \right),B\left( {{x}_{1}},{{y}_{1}} \right),C\left( 3,2 \right),D\left( {{x}_{2}},{{y}_{2}} \right)$

We know that,

Sides of a square are equal

$AB=BC$

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

The distance between $AB$ and $BC$ is given by,

$\sqrt{{{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}}=\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}}$

Squaring on both sides and solving them, we get,

${{x}^{2}}+2x+1+{{y}^{2}}-4y+4={{x}^{2}}+9-6x+{{y}^{2}}+4-4y$

$8x=8$

$x=1$

We know that,

All interior angles in a square are of angle $90{}^\circ $

Consider $\vartriangle ABC$,

By Pythagoras theorem,

${{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( AC \right)}^{2}}$

Distance formula is used to find the distance of $AB,BC,AC$ and substituting them in the Pythagoras theorem,

${{\left( \sqrt{{{\left( 1+1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}} \right)}^{2}}+{{\left( \sqrt{{{\left( 1-3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}} \right)}^{2}}={{\left( \sqrt{{{\left( 3+1 \right)}^{2}}+{{\left( 2-2 \right)}^{2}}} \right)}^{2}}$

$4+{{y}^{2}}+4-4y+4+{{y}^{2}}-4y+4=16$

$2{{y}^{2}}+16-8y=16$

$2{{y}^{2}}-8y=0$

$2y\left( y-4 \right)=0$

$y=0\text{ or }4$

$\therefore $The other two vertices are $B\left( 1,0 \right)$ and $D\left( 1,4 \right)$

5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are plotted on the boundary at a distance of $1\text{ m}$ from each other. There is a triangular grassy lawn in the plot as shown in the following figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

(i)  Taking $A$ as origin, find the coordinates of the vertices of the triangle

Ans:
Given that,

• Saplings are plotted $1\text{ m}$ from each other

• By taking $A$ as origin, To find the coordinates of the vertices of the triangle

• From the diagram, it can be observed that the coordinates are $P\left( 4,6 \right),Q\left( 3,2 \right),R\left( 6,5 \right)$

Area of the triangle $=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

Substitute the values of $P,Q,R$ in the above formula, we get,

Area of triangle $PQR=\frac{1}{2}\left[ 4\left( 2-5 \right)+3\left( 5-6 \right)+6\left( 6-2 \right) \right]$

$=\frac{1}{2}\left[ -12-3+24 \right]$

$=\frac{9}{2}$ square unit

$\therefore $The coordinates are $P\left( 4,6 \right),Q\left( 3,2 \right),R\left( 6,5 \right)$

(ii) What will be the coordinates of the vertices of $M$ $PQR$ if $C$ is the origin? Also calculate the areas of the triangles in these cases. What do you observe?

Ans:

Given that,

• Saplings are plotted $1\text{ m}$ from each other

• Taking $C$ as origin, $CB$ as $x$axis, $CD$ as $y$ axis

• From the diagram, it can be observed that the coordinates are $P\left( 4,6 \right),Q\left( 3,2 \right),R\left( 6,5 \right)$

Area of the triangle $=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

Substitute the value of $P,Q,R$ in the above formula, we get,

Area of the triangle $PQR=\frac{1}{2}\left[ 12\left( 6-3 \right)+13\left( 3-2 \right)+10\left( 2-6 \right) \right]$

$=\frac{1}{2}\left[ 36-13+40 \right]$

$=\frac{9}{2}$ square unit

$\therefore $The coordinates are $P\left( 4,6 \right),Q\left( 3,2 \right),R\left( 6,5 \right)$

The area of triangles are the same in both cases.

6. The vertices of a $\vartriangle ABC$ are $A\left( 4,6 \right),B\left( 1,5 \right)$ and $C\left( 7,2 \right)$. A line is drawn to intersect sides $AB$and $AC$ at $D$ and $E$respectively, such that$\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{4}$. Calculate the area of the $\vartriangle ADE$ and compare it with the area of $\vartriangle ABC$. (Recall the converse of basic proportionality theorem and Theorem 6.6 related to Ratio of areas of two similar triangles.)

Ans:

Given that,

• The vertices of $\vartriangle ABC$ are $A\left( 4,6 \right),B\left( 1,5 \right),C\left( 7,2 \right)$

• $\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{4}$

• To calculate the area of $\vartriangle ADE$ and compare it with $\vartriangle ABC$

From the diagram, we observe that $D$ and $E$are the mid-points of $AB$ and $AC$ respectively so that they divide the line segment $AB$ and $AC$ in the ratio of $1:3$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$D\left( x,y \right)=\left[ \frac{1\left( 1 \right)+3\left( 4 \right)}{1+3},\frac{1\left( 5 \right)+3\left( 6 \right)}{1+3} \right]$

$=\left[ \frac{1+12}{4},\frac{5+18}{4} \right]$

$=\left[ \frac{13}{4},\frac{23}{4} \right]$

$E\left( x,y \right)=\left[ \frac{1\left( 7 \right)+3\left( 4 \right)}{1+3},\frac{1\left( 2 \right)+3\left( 6 \right)}{1+3} \right]$

$=\left[ \frac{7+12}{4},\frac{2+18}{4} \right]$

$=\left[ \frac{19}{4},\frac{20}{4} \right]$

Area of the triangle $=\frac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

Area of $\vartriangle ADE=\frac{1}{2}\left[ 4\left( \frac{23}{4}-\frac{20}{4} \right)+\frac{13}{4}\left( \frac{20}{4}-6 \right)+\frac{19}{4}\left( 6-\frac{23}{4} \right) \right]$

$=\frac{1}{2}\left[ 2-\frac{13}{4}+\frac{19}{16} \right]$

$=\frac{1}{2}\left[ \frac{67-52}{16} \right]$

$=\frac{15}{32}$square unit.

Area of $\vartriangle ABC=\frac{1}{2}\left[ 4\left( 5-2 \right)+1\left( 2-6 \right)+7\left( 6-5 \right) \right]$

$=\frac{1}{2}\left( 12-4+7 \right)$

$=\frac{15}{2}$ square unit

Thus the ratio of areas of $\vartriangle ADE$ to $\vartriangle ABC$ is $1:16$

If a line segment in a triangle divides its two sides in the same ratio, then the line segment is parallel to the third side of the triangle. Thus these triangles are similar to each other.

And the ratio between the areas of two triangles will be equal to the square of the ratio between the sides of the triangle.

$\therefore $The ratio between the areas of the $\vartriangle ADE$ and $\vartriangle ABC$ is $1:16$

7. Let $A\left( 4,2 \right),B\left( 6,5 \right)$ and $C\left( 1,4 \right)$ be the vertices of the $\vartriangle ABC$

(i) The median from $A$ meets $BC$ at $D$. Find the coordinates of point $D$

Ans:

Given that,

$A\left( 4,2 \right),B\left( 6,5 \right),C\left( 1,4 \right)$ are the vertices of the given triangle.

To find,

The coordinates of point $D$

Let $AD$ be the median of the given triangle.

So median $D$ is the midpoint of $BC$

So, 

$D\left( x,y \right)=\left( \frac{6+1}{2},\frac{5+4}{2} \right)$

$=\left( \frac{7}{2},\frac{9}{2} \right)$

$\therefore $The coordinate of $D$ is $\left( \frac{7}{2},\frac{9}{2} \right)$

(ii) Find the coordinates of point $P$ on $AD$ such that $AP:PD=2:1$

Ans:

Given that,

$A\left( 4,2 \right),B\left( 6,5 \right),C\left( 1,4 \right)$ are the vertices of the given triangle.

$AP:PD=2:1$

To find,

The coordinate of $P$

Point $P$ divides $AB$ in the ratio of $m:n=2:1$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$P\left( x,y \right)=\left[ \frac{2\left( \frac{7}{2} \right)+1\left( 4 \right)}{2+1},\frac{2\left( \frac{9}{2} \right)+1\left( 2 \right)}{2+1} \right]$

$=\left[ \frac{7+4}{3},\frac{9+2}{3} \right]$

$=\left( \frac{11}{3},\frac{11}{3} \right)$

∴The coordinates of p is $P\left( \frac{11}{3},\frac{11}{3} \right)$.

(iii) Find the coordinate of point $Q$ and $R$ on medians $BE$ and $CF$ respectively such that $BQ:QE=2:1$ and $CR:RF=2:1$. What do you observe?

Ans:

Given that,

$A\left( 4,2 \right),B\left( 6,5 \right),C\left( 1,4 \right)$ are the vertices of the given triangle.

$BQ:QE=2:1$

$CR:RF=2:1$

To find,

Coordinate of $Q$ and $R$

4to find the coordinate of $E$

Median $BE$ of the triangle will divide the side $AC$ in two equal parts.

So $E$ is the midpoint of $AC$

$E=\left( \frac{4+1}{2},\frac{2+4}{2} \right)$

$=\left( \frac{5}{2},\frac{6}{2} \right)$

$=\left( \frac{5}{2},3 \right)$

$Q$ divides $BE$ in the ratio of  $2:1$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$Q\left( x,y \right)=\left[ \frac{2\left( \frac{5}{2} \right)+1\left( 6 \right)}{2+1},\frac{2\left( 3 \right)+1\left( 5 \right)}{2+1} \right]$

$=\left[ \frac{5+6}{3},\frac{6+5}{3} \right]$

$=\left( \frac{11}{3},\frac{11}{3} \right)$

Median $CF$ divide $AB$ in the ratio of $2:1$.

So $F$ is the midpoint of $AB$

$F=\left( \frac{4+6}{2},\frac{2+5}{2} \right)$

$=\left( \frac{10}{2},\frac{7}{2} \right)$

$=\left( 5,\frac{7}{2} \right)$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$R\left( x,y \right)=\left[ \frac{2\left( 5 \right)+1\left( 1 \right)}{2+1},\frac{2\left( \frac{7}{2} \right)+1\left( 4 \right)}{2+1} \right]$

$=\left[ \frac{10+1}{3},\frac{7+4}{3} \right]$

$=\left( \frac{11}{3},\frac{11}{3} \right)$

The coordinates of $Q$ and that of $R$ is $\left( \frac{11}{3},\frac{11}{3} \right)$

(iv) What do you observe?

Ans:

Given that,

$A\left( 4,2 \right),B\left( 6,5 \right),C\left( 1,4 \right)$ are the vertices of the given triangle.

From the above observation, we found that the coordinates of $P,Q,R$ are all the same. So, all these points are representing the same point on the same plane which is the centroid of the triangle.

(v) If $A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)$ and $C\left( {{x}_{3}},{{y}_{3}} \right)$ are the vertices of $\vartriangle ABC$, find the coordinates of the centroid of the triangle.

Ans:

Given that,

$A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)$and $C\left( {{x}_{3}},{{y}_{3}} \right)$ are the vertices of the $\vartriangle ABC$

To find,

The coordinate of the centroid of the triangle

Median $AD$ divides the side $BC$ into two equal parts. So $D$ is the midpoint of $BC$

$D=\left( \frac{{{x}_{2}}+{{x}_{3}}}{2},\frac{{{y}_{2}}+{{y}_{3}}}{2} \right)$

Let the centroid of this triangle be $O$ and $O$ divides the side $AD$ in the ratio of $2:1$

By section formula,

$P\left( x,y \right)=\left[ \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right]$

$O\left( x,y \right)=\left[ \frac{2\left( \frac{{{x}_{2}}+{{x}_{3}}}{2} \right)+1\left( {{x}_{1}} \right)}{2+1},\frac{2\left( \frac{{{y}_{2}}+{{y}_{3}}}{2} \right)+1\left( {{x}_{1}} \right)}{2+1} \right]$

$=\left[ \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right]$

The centroid of the triangle $ABC$ with vertices $A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)$,$C\left( {{x}_{3}},{{y}_{3}} \right)$ is $\left[ \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right]$

8. $ABCD$ is a rectangle formed by the points $A\left( -1,-1 \right),B\left( -1,4 \right),C\left( 5,4 \right)$ and $D\left( 5,-1 \right)$. $P,Q,R$ and $S$ are the midpoints of $AB,BC,CD$ and $DA$ respectively. Is the quadrilateral $PQRS$a square? A rectangle? Or a rhombus? Justify your answer.

Ans:

Given that,

• $A\left( -1,-1 \right),B\left( -1,4 \right),C\left( 5,4 \right),D\left( 5,-1 \right)$ are the vertices of the rectangle.

To find,

If $PQRS$ is a square or rectangle or rhombus

Form the figure,

$P$ is the midpoint of $AB$

$P=\left( \frac{-1-1}{2},\frac{-1+4}{2} \right)$

$=\left( -\frac{2}{2},\frac{3}{2} \right)$

$=\left( -1,\frac{3}{2} \right)$

$Q=\left( \frac{-1+5}{2},\frac{4+4}{2} \right)$

$=\left( 2,4 \right)$

$R=\left( \frac{5+5}{2},\frac{4+\left( -1 \right)}{2} \right)$

$=\left( 5,\frac{3}{2} \right)$

$S=\left( \frac{-1+5}{2},\frac{-1-1}{2} \right)$

$=\left( 2,-1 \right)$

The distance between any two points is given by the Distance formula,

$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$

Distance between the points $P$ and $Q$ is,

$PQ=\sqrt{{{\left( -1-2 \right)}^{2}}+{{\left( \frac{3}{2}-4 \right)}^{2}}}$

$=\sqrt{9+\frac{25}{4}}$

$=\sqrt{\frac{61}{4}}$

Distance between the points $Q$ and $R$ is,

$QR=\sqrt{{{\left( 2-5 \right)}^{2}}+{{\left( 4-\frac{3}{2} \right)}^{2}}}$

$=\sqrt{9+\frac{25}{4}}$

$=\sqrt{\frac{61}{4}}$

Distance between the points $R$ and $S$ is,

$RS=\sqrt{{{\left( 5-2 \right)}^{2}}+{{\left( \frac{3}{2}+1 \right)}^{2}}}$

$=\sqrt{9+\frac{25}{4}}$

$=\sqrt{\frac{61}{4}}$

Distance between the points $S$ and $P$ is,

$SP=\sqrt{{{\left( 2+1 \right)}^{2}}+{{\left( -1-\frac{3}{2} \right)}^{2}}}$

$=\sqrt{9+\frac{25}{4}}$

$=\sqrt{\frac{61}{4}}$

Distance between the diagonals $P$ and $R$ is,

$PR=\sqrt{{{\left( -1-5 \right)}^{2}}+{{\left( \frac{3}{2}-\frac{3}{2} \right)}^{2}}}$

$=\sqrt{{{6}^{2}}}$

$=6$

Distance between the diagonals $Q$ and $S$ is,

$QS=\sqrt{{{\left( 2-2 \right)}^{2}}+{{\left( 4+1 \right)}^{2}}}$

$=\sqrt{{{5}^{2}}}$

$=5$

From the above calculation, we observe that all sides of the quadrilateral are of the same length but the diagonals are different lengths. 

 $\therefore PQRS$ is a rhombus..

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry - PDF Download

You can opt for Chapter 7 - Coordinate Geometry NCERT Solutions for Class 10 Maths PDF for Upcoming Exams, and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 10 Maths PDFs (Chapter-wise)

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

Coordinate Geometry

Coordinate Geometry is the branch of Mathematics that helps us exactly locate a given point with the help of an ordered pair of numbers. Coordinate geometry is also known as Cartesian Geometry. It helps you find the distance between two points whose coordinates are given. You can also find the coordinates of the point which divides the line segment joining two given points in the given ratio. Also you will learn how to find the area of a triangle in terms of the coordinate of its vertices.

There are five sections and four exercises covered under Class 10 Maths Chapter 7 NCERT Solutions. The five sections are introduction to coordinate geometry, formula to calculate distance between two pints, finding the coordinate of the point dividing the line in a particular ratio called section formula and finding the area of triangle in the form of coordinates of their vertices.

And the Related Four Exercises Based on These Concepts are:

• Exercise 7.1 Introduction and  Distance Formula

• Exercise 7.2 Section Formula

• Exercise 7.3 area of Triangle

• Exercise 7.4 Miscellaneous examples

Exercise 7.1 consisted of the basic Coordinate Geometry problems and problems on distance between two points. It helps you to find the distance between any two points given on a plane.

Exercise 7.2 consisted of Section Formula. Section Formula deals with finding the coordinates of a point dividing the line in the particular ratio.

Exercise 7.3 has problems on the area of the triangle in the form of coordinates of the vertices.

Exercise 7.4 has then miscellaneous problems on Distance Formula, Section Formula and Area of Triangle.

NCERT Solutions for Class 10 Maths Chapter 7 Excercises

What is Coordinate Geometry

We have studied number lines to mark any point. Descartes invented the idea of placing two perpendicular lines to each other on a plane, and locating points on the lines of the plane. The perpendicular lines may be in any direction  but usually one line is horizontal and the other vertical.

Coordinate Geometry is the branch of mathematics that helps us to exactly locate a given point with the help of an ordered pair of numbers. Coordinate geometry is the combination of geometry and algebra to solve the problems.

Coordinate geometry is also said to be the study of graphs. Graphs are visual representations of our data. It can be in different forms like bar graphs, histograms, line graphs, etc.

In these NCERT Solutions for Chapter 7 Maths Class 10, students will study different concepts and formulas related to coordinate geometry.

Terms Related to Coordinate Geometry

While studying coordinate geometry the students must be aware of some important terms related to coordinate geometry they are well explained in this chapter.

From this figure, let us understand some important terms used in the coordinate formula of geometry.

Axes of Coordinates

In the above figure OX and OY are called X-axis and Y-axis respectively and both together are known as axes of coordinates.

Origin

The point of intersection of the axes is called the origin; it is O.

Abscissa

The distance of any point on the plane from the Y-axis is called the abscissa.

Ordinate

The distance of any point on the plane from the X-axis is said to be ordinate.

Coordinate of the Origin

It has zero distance from both the axes. Therefore the coordinates of the origin are (0, 0).

Quadrant

The axes divide the plane into four parts. These four parts are said to be quadrants.

A quadrant is ¼ th part of a plane divided by coordinate axes. The plane is called the coordinate plane or the XY-plane and the axes are called the coordinate axes.

In the first quadrant, both the coordinates are positive.

In the second quadrant, the y-coordinate is positive and x-coordinate is negative.

In the third quadrant, both the coordinates are negative.

In the fourth quadrant, the y-coordinate is negative and the x-coordinate is positive.

Distance Formula

In chapter 7 Maths Class 10 we learn about distance formulas for finding distance between two points. Finding the distance between the two points by using formula when the two coordinates of the points are given.

Distance between two points O(x1,y1) and B(x2, y2) =OB = \[\sqrt{[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]}\]

Section Formula

Section formula helps us in finding the coordinates of the point dividing the line in the ratio m:n.

If P is the point dividing the line AB in the ratio m:n where coordinates of A(x1, y1) and B(x2, y2).

The coordinates of P will be ( mx2 + nx1/ m+ n , my2 + ny1/ m+ n)

Area of Triangle

It will let us find the area of any triangle in terms of coordinates of its vertices. This formula will also be used in finding the area of quadrilaterals.

Area of a Triangle = ½ |x1(y2−y3)+x2(y3–y1)+x3(y1–y2)|

All the basic formulas of coordinate geometry will help you to solve your geometry and algebra problems. 

Chapter 7 maths Class 10 provides you the list of all formulas of coordinate geometry to solve all the related problems.

Table of All Formulas of Coordinate Geometry

It is very important to get acquainted with these basic formulas of coordinate geometry. Chapter 7 Maths Class 10 coordinate formulas have been proven very helpful for students to solve the problems at a competitive level.

Historical Facts

• Rene Descartes, the Great French Mathematician of the seventeenth century, liked to lie in bed and think. One day, when lying in the bed  he solved the problem of describing the position of a point in a plane. His method was based on the old concept of latitude and longitude

• Rene Descartes(1596-1650), a French Mathematician, came up with a system known as the cartesian coordinate system to describe the positions of points and lines in a plane. 

• His latin name was Renatius Cartesius, hence the name cartesian plane was derived. He was referred to as the father of analytical geometry.

• Through cartesian coordinate or rectangular coordinate system, we can study the problems involving both geometry and algebra,

• Cartesian geometry is also termed as analytic geometry.

Regular Practice Is The Key To Success.

“Success isn’t something that just happens, success is learned, success is practiced.”

                                                                                                            -Sparky Anderson

Today it is strongly believed that success comes from regular practice. Intelligence also sometimes fails if not practiced. Maths needs a regular practice for perfection. Some people think Maths as a very tough subject, but if they have a regular practice of solving the problems they can crack any problem. Regular practice reduces the silly mistakes done during the examinations. Practising lots of problems makes you aware of all the possible questions related to any topic. It makes it easier to solve any problem during your final exams.

NCERT syllabus is designed according to the caliber of the students. Practicing NCERT problems after understanding the concept makes the students more confident in that area. Practicing NCERT solutions brings more accuracy and increases the speed of problem-solving skills. Practicing not only improves your conceptual understanding but also improves your logical reasoning.

Understanding the concept thoroughly and then solving the NCERT question will definitely increase your score in examinations.

NCERT Solutions Provide the Following Benefits

• Improves your accuracy

• Makes the concepts more clear

• Increases your speed.

• It improves your logical reasoning.

• It makes you more confident in a particular area.

What is Special About Vedantu’s NCERT Solutions?

Vedantu is a team of expert teaching professionals. Vedantu always tries to make the learning of any concept at the student level easy to understand. It explains the concept stepwise and more precisely along with the pictorial representation of the concept. The NCERT solutions are cross-checked by expert Mathematicians. Vedantu is a strong believer to impart quality education to students. NCERT solutions are given in simple language with alternate solutions and diagrammatic representation for easy understanding of the problem. Along with these Vedantu tries to make the learning fun by lowering the pressure of studies.

The Main Features of Vedantu’s NCERT Solution for Class 10 Coordinate Geometry are as Follows:

1. Vedantu’s NCERT solution for Class 10 coordinate geometry is designed in a very simple language.

2. The solutions are divided into parts so that they can be easily understood by the students.

3. The explanation is given in stepwise format.

4. All the solutions are given at the student's level of understanding.

5. The solutions are precise, brief and with relevant diagrams.

6. They are given in proper stepwise format along with the necessary comments.

7. The solutions are based on the latest NCERT syllabus and modern exam specifications.

8. The solutions are properly formatted to reduce the unnecessary burden of lengthy solutions.

Vedantu is a No.1 online teaching app. It provides you free pdf for Coordinate Geometry Class 10 NCERT Solutions. It also has the potential to help you score more in the examination. Vedantu tries their level best to provide quality education, by providing NCERT solutions for coordinate geometry Class 10. It also provides more practice questions for bringing the perfection in the concepts. NCERT solutions are 100% accurate. And they will be very helpful in the examination. Keep this Vedantu NCERT solution handy so that there will be no inconvenience during your study of coordinate geometry. NCERT solutions for Class 10 will help you to increase your score in the final examination.

Why is Coordinate Geometry Important?

Coordinate geometry helps you to locate the exact location of a point in a plane. Coordinate geometry provides us the link between algebra and geometry through lines, graphs, curves and equations.

Coordinate geometry is also said to be the study of graphs. Graphs are visual representations of our data. It can be in different forms like bar graphs, histograms, line graphs, etc. 

Coordinate geometry Class 10 hlds a weightage of total 6 marks in the final board examination.Coordinate geometry basically explains four concepts introduction to coordinate geometry, distance formula, section formula and area of triangle.

Deeper into Exercises

Each exercise is solved by vedantu experts. Every concept is followed by well defined exercises.Exercises are aimed to test your conceptual knowledge. All the problems of the exercises are based on the concepts explained earlier. Every problem has a step wise explanation. Students can easily understand the concept used behind the problems.besides these problems more examples are given to solve.

Different types of problems are given to understand the concept thoroughly.

Vedantu has studied all the exercise and variety of problems involved in each exercise.

Exercise 7.1

Exercise one is based on how to find distance between two points given on a plane.  Exercise 7.1 contains a total 10 questions each of different types.

By the end of this exercise the student will be able to find the distance between two points. With this property there are different types of problems in the exercise 7.1.

We can calculate the distance between points and prove the types of triangle are isosceles, equilateral or types  quadrilateral square, rhombus, rectangle or  whether the points are collinear or not.

Vedantu provides step wise solutions for every problem with relevant diagrams. Diagrams help us visualize the problem. With the distance formula we can even find the unknown values of x and y coordinate if the distance between the two points are given. Vedantu provides a stepwise explanation  of the solution so that you can solve any similar problems..

Exercise 7.2

Exercise 7.2 deals with the section formula that is finding the coordinates of the point which divides the given line any any specific ratio. Section formula is derived by a well defined theorem. Exercise 7.2 has a total of 10 questions.It is basically given on four types of problems.

Type1: Finding the section ratio or the end points of the segment when the section point is given.

Type 2: Finding the section point when the section ratio is given.

Type 3: Determination of the type of a given quadrilateral.

Type 4: Finding the unknown vertex from given points. 

Every problem is based on distance formula and section formula. The solutions are given step wise with the conceptual explanation. Which will make the concepts more clear.

Exercise 7.3

Earlier we calculated the area of the triangle using formula. In these chapter 7 Coordinate geometry students have to find the area of the triangle in terms of the coordinates of its vertices.

We can find the area of a triangle by finding the sides of the triangle by distance formula and then finding the area of the triangle by Heron's formula. But it becomes tedious if the length of the sides are in the form of irrational numbers. So we prefer to calculate the area in the terms of the coordinates of the vertices of the triangle. It will also help us to find the type of polygon when the coordinates are given.

Using the formula for the area of the triangle we can even calculate the area of the quadrilateral, as the quadrilateral breaks up into two triangles.

This exercise covers a total of 10 questions. They are based on four different types. 

Type 1: Finding the area of a triangle when the coordinates of the vertices are given.

Type 2: finding the area of a quadrilateral when coordinates of its vertices are given

Type 3: Finding collinearity of three points

Type 4: finding the desired results when the three points are collinear.

It has a total of 5 questions.

Exercise 7.4

Exercise 7.4 is based on the miscellaneous concepts which is covered that is distance formula, section formula and area of triangle. There are a total of 8 questions. Which will be helpful for you to check your knowledge regarding the concepts. Exercise 7 is the test of your knowledge on overall coordinate geometry, students study the concepts and solve therealted problems, but when all the problems related to different concepts are given they get confused. Hence Exercise 7.4 is given to check your knowledge for the concepts are clear or not. All the 8 questions are of different types.

Solving all these exercises will help you to master the coordinate geometry for Class 10. Vedantu’s NCERT solutions will prove very helpful in solving these exercises with more ease.

Summary

• To locate the position of an object or a point in a plane we require two perpendicular lines. One of them is horizontal and the other is vertical.

• The plane is called the cartesian plane or the coordinate plane and the lines are called the coordinate axes.

• The horizontal line is called the X-axis and the vertical line is called the Y-axis.

• The abscissa and ordinate of a given point are the distances of the point from Y-axis and X-axis respectively.

• The coordinates of any point on x-axis are of the form(x , 0).

• The coordinates of any point on x-axis are of the form(0, y).

• The distance between points P(x1 , y1) and Q( x2 , y2) is given by PQ = √[(x2 − x1)2 + (y2 − y1)2]

• Distance of point P( x, y) from the origin(0, 0) is given by OP = (x2 + y2)

• The coordinates of the point which divides the line joining the points P( x1, y1) and Q( x2, y2) internally in the ration m : n The coordinates of point will be ( mx2 + nx1/ m+ n , my2 + ny1/ m+ n)

• The coordinates of the mid-point of the line segment joining the points P( x1, y1) and Q( x2, y2) .

• The coordinates of point will be ( x1 + x2/2 , y1 + y2/ 2) 

• The coordinates of the centroid of triangle formed by the points A(x1 , y1) , B( x2, y2) and C(x3, y3) are ( x1 + x2 + x3/3 , y1 + y2 + y3/ 3)

• The area of the triangles formed by the points A(x1 , y1) , B( x2, y2) and C(x3, y3) is Area of a Triangle = ½ |x1(y2−y3)+x2(y3–y1)+x3(y1–y2)|

• If points A(x1 , y1) , B( x2, y2) and C(x3, y3) are collinear, then x1(y2−y3)+x2(y3–y1)+x3(y1–y2) = 0