NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry - Exercise 7.4

Exercise 7.4

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

Ans: Given  the points $\text{A}\left( \text{2,-2} \right)$ and $\text{B}\left( \text{3,7} \right)$

Given  the line $\text{2x+y-4=0}$.

• Let the given line $\text{2x+y-4=0}$ divide the line segment joining the points $\text{A}\left( \text{2,-2} \right)$  and $\text{B}\left( \text{3,7} \right)$  in a ratio $\text{k:1}$  at the point $\text{C}$ .

(Image will be uploaded soon)

By section formula, we have,

$\text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right)$  , where  $\text{P}\left( \text{x,y} \right)$ divides the join of$\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right)$ and $\left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right)$  in the ratio $\text{m:n}$. 

Therefore following the formula Coordinates of the point of division ,$\text{C}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{3k+2}}{\text{k+1}}\text{,}\dfrac{\text{7k-2}}{\text{k+1}} \right)$

This point of division $\text{C}$also lies on the line$\text{2x+y-4=0}$……(1)

By substituting the coordinates value of $C\left( x,y \right)$ in equation(1), we get,

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

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

By cross multiplication and further computing, 

$\Rightarrow 9k-2=0$

$\Rightarrow k=\dfrac{2}{9}$

Therefore, the line $\text{2x+y-4=0}$ divides the line segment joining the points $\text{A}\left( \text{2,-2} \right)$  and $\text{B}\left( \text{3,7} \right)$ in the ratio$2:9$ internally.

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

Ans: Given the points$\left( \text{x,y} \right)$,$\left( \text{1,2} \right)$and$\left( \text{7,0} \right)$. Let us denote the given points by $\text{A,B,C}$ respectively. Then ,

• Let $\text{A}\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)\text{=}\left( \text{x,y} \right)$

• Let $\text{B}\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)\text{=}\left( \text{1,2} \right)$

• Let  $\text{C}\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)\text{=}\left( \text{7,0} \right)$

Consider the formula for area of a triangle whose vertices are $\left( {{x}_{1}},{{y}_{1}} \right)$ ,$\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)$ , we have,

Area of a triangle$\text{=}\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\}$

For collinear points, area of the triangle formed by those points is $\text{0}$ .

Therefore, we have,

 for collinear points,   $\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\}\text{=0}$   …….(1)  

 By substituting the values of vertices from the given points$\text{A,B,C}$ in the Equation (1), we get,

Area of $\vartriangle \text{ABC}$$\text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{x}\left( \text{2-0} \right)\text{+1}\left( \text{0-y} \right)\text{+7}\left( \text{y-2} \right) \right]$

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

Simplifying, we get,

$\dfrac{\text{1}}{\text{2}}\left[ \text{2x+6y-14} \right]\text{=0}$

$\Rightarrow \text{2x+6y-14=0}$

$\Rightarrow \text{x+3y-7=0}$

 Then the required relation between $x$  and $y$ is given by  $\text{x+3y-7=0}$ .

3. Find the centre of a circle passing through the points $\left( \text{6,-6} \right)$ ,$\left( \text{3,-7} \right)$ and $\left( \text{3,3} \right)$ .

Ans: Given the points $\left( \text{6,-6} \right)\text{,}\left( \text{3,-7} \right)$and $\left( \text{3,3} \right)$ passing through a circle. Let us denote the given points by $\text{A,B,C}$ respectively. Then ,

• Let $P\left( \text{x,y} \right)$be the centre of the circle passing through the points $\left( \text{6,-6} \right)\text{,}\left( \text{3,-7} \right)$and $\left( \text{3,3} \right)$.

• The points $\text{A}\left( \text{6,-6} \right)\text{,B}\left( \text{3,-7} \right)$and $\text{C}\left( \text{3,3} \right)$ are the points on the circumference of the circle.

(Image will be uploaded soon)

Now , consider the distance formula,

Distance between a pair of points is given by $\sqrt{{{\left( {{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}} \right)}^{\text{2}}}\text{+}{{\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right)}^{\text{2}}}}$ .

Distance from centre$P\left( \text{x,y} \right)$ to$A$ ,$B$ and $C$ are  as following, using the Distance formula,

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

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

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

From the figure that $\text{PA=PB}$, since  $\text{PA}$ and $\text{PB}$ are the radii of the same circle.

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

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

Simplifying,

$\text{-6x-2y+14=0}$

$\Rightarrow 3x+y=7$…… (1)

Similarly $PA=PC$ since $\text{PA}$ and $\text{PC}$ are the radii of same circle.

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

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

$\Rightarrow \text{-6x+18y+54=0}$

$\Rightarrow \text{-3x+9y=-27}$ ……(2)

On adding equation (1) and (2), we get,

$\text{10y=-20}$

$\text{y=-2}$

Substituting the value of $\text{y}$ in equation (1) , we obtain 

$\text{3x-2=7}$

$\text{3x=9}$

$\text{x=3}$

Therefore , the coordinates of the centre$P\left( \text{x,y} \right)$ of the circle is $\left( \text{3,-2} \right)$ .

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

Ans: Given two opposite vertices of a square$\left( \text{-1,2} \right)$and$\left( \text{3,2} \right)$. Let us denote the given points by $\text{A,C}$ respectively. Then ,

• Let $ABCD$  be a square having  vertices$\text{A}\left( \text{-1,2} \right)$ and $\text{C}\left( \text{3,2} \right)$.

• Let $B\left( x,y \right)$and $D\left( {{x}_{1}},{{y}_{1}} \right)$ be the unknown vertices.

(Image will be uploaded soon)

Since  the sides of a square are equal to each other.

We have, $\text{AB=BC}$

Now , consider the distance formula,

Distance between a pair of points is given by $\sqrt{{{\left( {{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}} \right)}^{\text{2}}}\text{+}{{\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right)}^{\text{2}}}}$ .

By  the distance formula the distance $\text{AB}$ is given by,

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

And the distance $\text{BC}$ is given by ,

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

Therefore the condition $\text{AB=BC}$ implies,

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

Squaring both sides and expanding,

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

$\Rightarrow 8x=8$

$\Rightarrow x=1$

Then coordinates of  $\text{B}$ is $\left( \text{1,y} \right)$ .

Now we know, all the interior angles are of ${{90}^{\circ }}$ in a square.

Then by Pythagoras theorem in the triangle $\vartriangle ABC$,

$A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$

Using distance formula, the distance $\text{AC}$ is given by

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

Therefore the condition $A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$ implies

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

Simplifying and expanding,

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

$\Rightarrow \text{2}{{\text{y}}^{\text{2}}}\text{+16-8y=16}$

$\Rightarrow \text{2}{{\text{y}}^{\text{2}}}\text{-8y=0}$

$\Rightarrow \text{y}\left( \text{y-4} \right)\text{=0}$

$\Rightarrow \text{y=0 or 4}$

Hence the required unknown verticles of the square $\text{ABCD}$ 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 planted on the boundary at a distance of $\text{1 m}$ 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.

(Image will be uploaded soon)

(i)Taking $\text{A}$  as origin, find the coordinates of the vertices of the triangle.

    Also calculate the area of the triangle in this case. 

Ans:

• If we take $\text{A}$  as origin, the line segment  $\text{AD}$  will represent the $\text{x-axis}$  and $AB$  will represent the $\text{y-axis}$.

• Here it is mentioned that saplings of gulmohar are planted $\text{1 m}$ apart from each other along $\text{AD}$ and  $AB$.

• Then  it is  obvious from figure that the coordinates of point $P$ , $Q$and $R$ are $\left( 4,6 \right)$ ,$\left( 3,2 \right)$ and $\left( 6,5 \right)$ respectively. Then we have,

$P\left( {{x}_{1}},{{y}_{1}} \right)=\left( 4,6 \right)$

$Q\left( {{x}_{2}},{{y}_{2}} \right)=\left( 3,2 \right)$

$R\left( {{x}_{3}},{{y}_{3}} \right)=\left( 6,5 \right)$

Consider the formula for area of a triangle whose vertices are $\left( {{x}_{1}},{{y}_{1}} \right)$ ,$\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)$ , we have,

Area of a triangle$\text{=}\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\}$ …… (1)

By substituting the values of vertices $P$ , $Q$and $R$obtained  in the Equation (1),

Area of $\vartriangle \text{PQR}$$\text{=}\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\}$

$\text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{4}\left( \text{2-5} \right)\text{+3}\left( \text{5-6} \right)\text{+6}\left( \text{6-2} \right) \right]$

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

$\text{=}\dfrac{\text{9}}{\text{2}}\text{ square units}$

(ii) What will be the coordinates of the vertices of $\vartriangle \text{PQR}$ if $\text{C}$  is the origin? Also calculate the areas of the triangle in this case. What do you observe?

Ans:

• If we take $C$ as origin, $\text{CB}$ will represent the $\text{x-axis}$ , and $\text{CD}$  will represent the $\text{y-axis}$ . 

• Here it is mentioned that saplings of gulmohar are planted $\text{1 m}$ apart from each other along $\text{CD}$ and  $\text{CB}$.

• Then it is obvious from the figure 5 that the coordinates of vertices $P$ , $Q$and $R$ are $\left( 12,2 \right)$ ,$\left( 13,6 \right)$and $\left( 10,3 \right)$ respectively. Then we have,$\text{P}\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)\text{=}\left( \text{12,2} \right)$

$\text{Q}\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)\text{=}\left( \text{13,6} \right)$

$\text{R}\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)\text{=}\left( \text{10,3} \right)$

Consider the formula for area of a triangle whose vertices are $\left( {{x}_{1}},{{y}_{1}} \right)$ ,$\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)$ , we have,

Area of a triangle$\text{=}\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\}$ …… (1)

By substituting the values of vertices$P$ , $Q$and $R$  obtained in the equation (1),

Area of triangle $\vartriangle \text{PQR}$$\text{=}\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\}$

$\text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{12}\left( \text{6-3} \right)\text{+13}\left( \text{3-2} \right)\text{+10}\left( \text{2-6} \right) \right]$

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

$\text{=}\dfrac{\text{9}}{\text{2}}\text{ square units}$

It can be observed that the area of the triangle is same in both the cases i.e taking $\text{A}$ as origin and taking $\text{C}$ as origin.

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

Ans: Given the points $\text{A}\left( \text{4,6} \right)$,$\text{B}\left( \text{1,5} \right)$and$\text{C}\left( \text{7,2} \right)$.

Given that $\dfrac{\text{AD}}{\text{AB}}\text{=}\dfrac{\text{AE}}{\text{AC}}\text{=}\dfrac{\text{1}}{\text{4}}$

(Image will be uploaded soon)

Therefore the points $\text{D}$ and $\text{E}$ divide $\text{AB}$ and $\text{AC}$ respectively  in the ratio $1:3$ .

By section formula, we have,

$\text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right)$  , where  $\text{P}\left( \text{x,y} \right)$ divides the join of$\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right)$ and $\left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right)$  in the ratio $\text{m:n}$. 

Therefore Coordinates of  the point $D=\left( \dfrac{1\times 1+3\times 4}{1+3},\dfrac{1\times 5+3\times 6}{1+3} \right)$

$=\left( \dfrac{13}{4},\dfrac{23}{4} \right)$

And Coordinates of the point $E=\left( \dfrac{1\times 7+3\times 4}{1+3},\dfrac{1\times 2+3\times 6}{1+3} \right)$

$=\left( \dfrac{19}{4},\dfrac{20}{4} \right)$

Consider the formula for area of a triangle whose vertices are $\left( {{x}_{1}},{{y}_{1}} \right)$ ,$\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)$ , we have,

Area of a triangle$\text{=}\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\}$ …… (1)

By substituting the vertices $\text{A,D,E}$ obtained in equation (1),

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

$\text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{3-}\dfrac{\text{13}}{\text{4}}\text{+}\dfrac{\text{19}}{\text{16}} \right]$

$\text{=}\dfrac{\text{15}}{\text{32}}\text{ }square\text{ unit}$

Substituting the vertices $\text{A,B,C}$ from the given points in equation (1), 

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

$=\dfrac{15}{2}$

Therefore , $\dfrac{\text{area of }\vartriangle \text{ADE}}{area\text{ of }\vartriangle \text{ABC}}=\dfrac{\dfrac{15}{32}}{\dfrac{15}{2}}$

$=\dfrac{1}{16}$

Therefore the area of  $\vartriangle \text{ABC}$ is $16$ times the area of  $\vartriangle \text{ADE}$.

Alternatively, if two sides of a triangle is divided by a line segment in the same ratio then the line segment is parallel to the third side of the triangle and the two triangles formed bt this line segment is similar to the original. Therefore the ratio between the areas of these two triangle will be the square of the ratio of the sides of these two triangles formed.

Hence the ratio of the area of  $\vartriangle \text{ADE}$ to the area of $\vartriangle \text{ABC}$ is ${{\left( \dfrac{1}{4} \right)}^{2}}=\dfrac{1}{16}$ .

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

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

(Image will be uploaded soon)

Ans: Given the points$\text{A}\left( \text{4,2} \right)$ ,$\text{B}\left( \text{6,5} \right)$ and$\text{C}\left( \text{1,4} \right)$are the vertices of a triangle.

$\text{A}\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)\text{=}\left( \text{4,2} \right)$

$\text{B}\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)\text{=}\left( \text{6,5} \right)$

$\text{C}\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)\text{=}\left( \text{1,4} \right)$

Here it is given that the median from $\text{A}$  meets $\text{BC}$ at $\text{D }$. Then  $\text{AD}$ is median of 

the triangle$~ABC$.   Hence $D$ is the midpoint of $\text{BC}$ .

By section formula, we have,

$\text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right)$  , where  $\text{P}\left( \text{x,y} \right)$ divides the join of$\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right)$ and $\left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right)$  in the ratio $\text{m:n}$. 

Now  $D$divides  the line segment $\text{BC}$ in the ratio $\text{1:1}$

Therefore Coordinates of $D$$\text{=}\left( \dfrac{\text{6+1}}{\text{2}}\text{,}\dfrac{\text{5+4}}{\text{2}} \right)\text{=}\left( \dfrac{\text{7}}{\text{2}}\text{,}\dfrac{\text{9}}{\text{2}} \right)$

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

Ans: Given that $\text{AP:PD=2:1}$

 Hence the Point $\text{P}$  divides the side $AD$ in a ratio$2:1$.

By section formula, we have,

$\text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right)$  , where  $\text{P}\left( \text{x,y} \right)$ divides the join of$\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right)$ and $\left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right)$  in the ratio $\text{m:n}$. 

Therefore Coordinates of $\text{P}$is given by,

$\text{P=}\left( \dfrac{\text{2 }\!\!\times\!\!\text{ }\dfrac{\text{7}}{\text{2}}\text{+1 }\!\!\times\!\!\text{ 4}}{\text{2+1}}\text{,}\dfrac{\text{2 }\!\!\times\!\!\text{ }\dfrac{\text{9}}{\text{2}}\text{+1 }\!\!\times\!\!\text{ 2}}{\text{2+1}} \right)\text{=}\left( \dfrac{\text{11}}{\text{3}}\text{,}\dfrac{\text{11}}{\text{3}} \right)$

(iii) Find the coordinates of point $\text{Q}$  and $\text{R}$  on medians $\text{BE}$  and $\text{CF}$  respectively such that $\text{BQ: QE = 2:1}$ and $\text{CR: RF = 2:1}\text{.}$

Ans: Since  $\text{BE}$is the median of the triangle$~ABC$,$E$ is the mid-point of side $AC$ . Then $\text{BE}$ divides the side $AC$in the ratio $\text{1:1}$.

By section formula, we have,

$\text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right)$  , where  $\text{P}\left( \text{x,y} \right)$ divides the join of$\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right)$ and $\left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right)$  in the ratio $\text{m:n}$. 

Hence ,Coordinates of $E$ is given by,

$E=\left( \dfrac{4+1}{2},\dfrac{2+4}{2} \right)=\left( \dfrac{5}{2},3 \right)$

Now  given that $\text{BQ: QE = 2:1}$

Then thepoint $Q$ divides the side $BE$ in a ratio$2:1$ .

Hence the Coordinates $Q$of is given by,

$Q=\left( \dfrac{2\times \dfrac{5}{2}+1\times 6}{2+1},\dfrac{2\times 3+1\times 5}{2+1} \right)=\left( \dfrac{11}{3},\dfrac{11}{3} \right)$

Since $CF$is the median of the triangle $~ABC$ ,$\text{F}$  is the mid-point of side$AB$. Hence , $CF$will divide the side $AB$ in the ratio $\text{1:1}$.

Therefore Coordinates of $\text{F}$ is given by,

$F=\left( \dfrac{4+6}{2},\dfrac{2+5}{2} \right)=\left( 5,\dfrac{7}{2} \right)$

Now  given that $\text{CR: RF = 2:1}\text{.}$

Then the point $\text{R}$  divides the side $\text{CF}$  in a ratio $\text{2:1}$ .

Then coordinates of $\text{R}$is given by,

$R=\left( \dfrac{2\times 5+1\times 1}{2+1},\dfrac{2\times \dfrac{7}{2}+1\times 4}{2+1} \right)=\left( \dfrac{11}{3},\dfrac{11}{3} \right)$

(iv) What do you observe?

Ans: From the obtained values of $\text{P,Q,R}$it can be observed that, the coordinates of the points $\text{P,Q,R}$ are the same. Therefore, each of these points represent the same point on the plane i.e., the centroid of the triangle.

(v) If$\text{A(x1, y1)}$ , $\text{B}\left( \text{x2, y2} \right)$ and $\text{C}\left( \text{x3, y3} \right)$ are the vertices of$\vartriangle \text{ABC}$ , find the coordinates of the centroid of the triangle.

Ans: Given that$A\left( {{x}_{1}},\text{ }{{y}_{1}} \right)$ , $B\left( {{x}_{1}},{{y}_{2}} \right)$ and $C\left( {{x}_{3}},{{y}_{3}} \right)$ are the vertices of the triangle $\text{ABC}$ .The median $AD$ of the triangle divides the side $BC$ in the ratio $\text{1:1}$. Therefore, $D$  is the mid-point of side $BC$ .

By section formula, we have,

$\text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right)$  , where  $\text{P}\left( \text{x,y} \right)$ divides the join of$\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right)$ and $\left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right)$  in the ratio $\text{m:n}$. 

Then Coordinates of$D$ is given by, 

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

Let the centroid of the triangle $\text{ABC}$be $O$ .

Then the  Point $O$  divides the side $AD$ in the ratio $2:1$.

Then by the section formula, coordinates of  $O$is given by,  

$O=\left( \dfrac{2\times \dfrac{{{x}_{2}}+{{x}_{3}}}{2}+1\times {{x}_{1}}}{2+1},\dfrac{2\times \dfrac{{{y}_{2}}+{{y}_{3}}}{2}+1\times {{y}_{1}}}{2+1} \right)$

$=\left(\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$.

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

Ans: Given the points$\text{A}\left( \text{-1,-1} \right)$ ,$\text{B}\left( \text{-1,4} \right)$ ,$\text{C}\left( \text{5,4} \right)$,and $D\left( \text{5,-1} \right)$ are the vertices of a rectangle.

Given $P$ is the midpoint of $AB$. 

By section formula, we have,

$\text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right)$  , where  $\text{P}\left( \text{x,y} \right)$ divides the join of$\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right)$ and $\left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right)$  in the ratio $\text{m:n}$. 

Now $P$divides  the line segment $AB$ in the ratio $\text{1:1}$.

Therefore , the coordinates  of $P$ are $\left( \dfrac{\text{-1-1}}{\text{2}}\text{,}\dfrac{\text{-1+4}}{\text{2}} \right)\text{=}\left( \text{-1,}\dfrac{\text{3}}{\text{2}} \right)$

Similarly, the coordinates of $\text{Q}$,$R$ and $S$ are $\left( 2,4 \right)$ ,$\left( \text{5,}\dfrac{\text{3}}{\text{2}} \right)$ and $\left( 2,-1 \right)$respectively.

Now , consider the distance formula,

Distance between a pair of points is given by $\sqrt{{{\left( {{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}} \right)}^{\text{2}}}\text{+}{{\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right)}^{\text{2}}}}$ where $\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)$ and $\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)$ are the coordinates of the given pair of points.

Distance between the pair of  points $P$ and $Q$ is given by,

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

$=\sqrt{\text{9+}\dfrac{\text{25}}{\text{4}}}$

$=\sqrt{\dfrac{\text{61}}{\text{4}}}$

Distance between the pair of  points $Q$ and $R$ is given by,

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

$\text{=}\sqrt{\text{9+}\dfrac{\text{25}}{\text{4}}}$

$\text{=}\sqrt{\dfrac{\text{61}}{\text{4}}}$

Distance between the pair of  points $R$ and $S$ is given by,

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

$\text{=}\sqrt{\text{9+}\dfrac{\text{25}}{\text{4}}}$

$\text{=}\sqrt{\dfrac{\text{61}}{\text{4}}}$

Distance between the pair of points$S$ and $P$ is given by,

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

$\text{=}\sqrt{\text{9+}\dfrac{\text{25}}{\text{4}}}$

$\text{=}\sqrt{\dfrac{\text{61}}{\text{4}}}$

Distance between the pair of points $P$ and $R$ is given by,

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

$\text{=6}$

Distance between the pair of points $Q$ and $S$ is given by,

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

$\text{=5}$

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It can be observed that all sides $\text{PQ,SP,QR,RS}$  of the given quadrilateral $\text{PQRS}$  are of the same length but the diagonals$\text{PR,QS}$ are of different lengths. Therefore, $\text{PQRS}$ is a rhombus.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.4

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