NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.3) Exercise 7.3

EXERCISE NO: 7.3

1. Find the area of the triangle whose vertices are: \[\text{(2, 3), (- 1, 0), (2, - 4) }\]

Ans: Assume A\[({{x}_{1}},\text{ }{{y}_{1}})=\left( 2,\text{ }3 \right)\] , B\[\left( {{x}_{2}},\text{ }{{y}_{2}} \right)=\left( -1\text{ },\text{ }0 \right)\] , C\[\left( {{x}_{3}},\text{ }{{y}_{3}} \right)=\left( 2,-4 \right)\]

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

By putting the above values of A,B,C and D

We get,

=\[\frac{1}{2}\left\{ 2(0-(-4))+(-1)(-4-3)+2(3-0) \right\}\]

=\[\frac{1}{2}\left\{ 8+7+6 \right\}\]

=\[\frac{21}{2}\] square units

2. Find the area of the triangle whose vertices are: \[\text{(-5, -1), (3, -5), (5, 2) }\]

Ans: Assume A\[({{x}_{1}},\text{ }{{y}_{1}})=\left( -5,-1 \right)\],B\[\left( {{x}_{2}},\text{ }{{y}_{2}} \right)=\left( \text{3,-5} \right)\] , C\[\left( {{x}_{3}},\text{ }{{y}_{3}} \right)=\left( 5,2 \right)\]

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

By putting the above values of A,B,C and D

We get,

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

=\[\frac{1}{2}\left\{ 35+9+20 \right\}\]

=\[32\]square units

3 (i) In each of the following find the value of ‘k’, for which the points are collinear:\[\text{(7, - 2), (5, 1), (3, - k})\]

Ans: Assume A\[({{x}_{1}},\text{ }{{y}_{1}})=\left( 7,-2 \right)\],B\[\left( {{x}_{2}},\text{ }{{y}_{2}} \right)=\left( 5,1 \right)\] , C\[\left( {{x}_{3}},\text{ }{{y}_{3}} \right)=\left( 3,-k \right)\]

Area of a triangle for collinear point\[\frac{1}{2}\left\{ {{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right\}=0\]

By putting the above values of A,B,C and D

We get,

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

=\[7+7k-5k+10-9=0\]

=\[2k+8=0\]

=\[k=-4\]

As a result, the supplied points for \[k=-4\] are collinear.

3 (ii) In each of the following find the value of ‘k’, for which the points are collinear:\[\text{(8, 1), (k, - 4), (2, - 5)}\]

Ans: Assume A\[({{x}_{1}},\text{ }{{y}_{1}})=\left( 8,1 \right)\],B\[\left( {{x}_{2}},\text{ }{{y}_{2}} \right)=\left( k,-4 \right)\] , C\[\left( {{x}_{3}},\text{ }{{y}_{3}} \right)=\left( 2,-5 \right)\]

Area of a triangle for collinear point\[\frac{1}{2}\left\{ {{x}_{1}}({{y}_{2}}-{{y}_{1}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}}) \right\}=0\]

By putting the above values of A,B,C and D

We get,

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

=\[8-6k+10=0\]

=\[6k=18\]

=\[k=3\]

As a result, the supplied points for \[k=3\] are collinear.

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

Ans:

Assume A\[({{x}_{1}},{{y}_{1}})=(0,-1)\],B\[\left( {{x}_{2}},\text{ }{{y}_{2}} \right)=\left( 2,1 \right)\] , C\[\left( {{x}_{3}},\text{ }{{y}_{3}} \right)=\left( 0,3 \right)\]

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

Let D, E, and F be the midpoints of the triangle's sides. D, E, and F coordinates are determined by :

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

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

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

By putting the above values we get area of triangle DEF

=\[\frac{1}{2}\left\{ (2-1)+1(1-0)+0(0-2) \right\}\]

=\[\frac{1}{2}(1+1)=1\]Square units

Similarly,

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

=\[\frac{1}{2}(8)=4\]Square units.

As a result, Ratio of this area ∆DEF to the area of the triangle ∆ABC= \[1:4\]

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

Ans:

Let the quadrilateral's vertices be A $(4,2)$ , B$(3,5)$ , C $(3,-2)$, and D $(3,2)$.

Form two triangles, ABC and ACD, by joining AC.

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

By putting the values of A,B,C in the above formula we get:

=\[\frac{1}{2}\left\{ (-4)((-5)-(-2)+(-3)((-2)-(-2))+3((-2)-(-2) \right\}\]

=\[\frac{1}{2}(12+0+9)=\frac{21}{2}\]Square units

By putting the values of A,C,D in the above formula we get:

=\[\frac{1}{2}\left\{ (-4)((-2)-(-3)+3(3-(-2))+2((-2)-(-2) \right\}\]

=\[\frac{1}{2}(20+15+0)=\frac{35}{2}\]Square units

So, Area of  triangle ABCD= Area of triangle ABC + Area of triangle ACD$$

=$\left( \frac{21}{2}+\frac{35}{2} \right)=28$ Square units

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

Ans:

The triangle's vertices should be A (4, 6), B (3, 2), and C (5, 2).

Assume D is the midpoint of ABC's side BC.

As a result, in ABC, AD is the median.

D point coordinates are:$\left( \frac{3+5}{2}+\frac{-2+2}{2} \right)=(4,0)$

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

By putting the values of A,B,D in the above formula we get:

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

=\[\frac{1}{2}(-8+18-16)=-3\]Square units

By putting the values of A,B,D in the above formula we get:

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

=\[\frac{1}{2}(-8+32-30)=-3\]Square units

As we know area cannot be in negative. So, Area of triangle ADC is 3 Square units.

As a result, median AD has clearly separated triangle ABC into two equal-sized triangles.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.3

Opting for the NCERT solutions for Ex 7.3 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.3 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 7 Exercise 7.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 10 Maths Chapter 7 Exercise 7.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 10 Maths Chapter 7 Exercise 7.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.