NCERT Solutions class 10 Maths Exercise 7.3 Ch 7 Coordinate Geometry

NCERT Solutions for Class 10 Maths Exercise 7.3 Chapter 7 Coordinate Geometry – FREE PDF Download

NCERT Class 10 Maths Ch 7 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter
7 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Coordinate Geometry solutions will help you understand the chapter thoroughly.

NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry

1. Find the area of the triangle whose vertices are:
(i) (2, 3), (–1, 0), (2, –4)
(ii) (–5, –1), (3, –5), (5, 2)
Ans. (i) (2, 3), (–1, 0), (2, –4)
Area of Triangle =
[2 {0 − (−4)} – 1 (−4 − 3) + 2 (3 − 0)]
[2 (0 + 4) – 1 (−7) + 2 (3)]
(8 + 7 + 6) =  sq. units
(ii) (–5, –1), (3, –5), (5, 2)
Area of Triangle =
=  [−5 (−5 − 2) + 3 {2 − (−1)} + 5 {−1 − (−5)}]
[−5 (−7) + 3 (3) + 5 (4)]
(35 + 9 + 20)
(64) = 32 sq. units

2. In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, –2), (5, 1), (3, k)
(ii) (8, 1), (k, –4), (2, –5)
Ans. (i) (7, –2), (5, 1), (3, k)
Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.
Area of Triangle =
⇒   [7 (1 − k) + 5 {k − (−2)} + 3 (−2 − 1)] = 0
(7 − 7k + 5k + 10 − 9) = 0
⇒   (7 − 7k + 5k + 1) = 0
⇒   (8 − 2k) = 0
⇒ 8 − 2k = 0
⇒ 2k = 8
⇒ k = 4
(ii) (8, 1), (k, –4), (2, –5)
Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.
Area of Triangle =
⇒   [8 {−4 − (−5)} + k (−5 − 1) + 2 {1 − (−4)}] = 0
(8 − 6k + 10) = 0
⇒  (18 − 6k) = 0
⇒18 − 6k = 0
⇒ 18 = 6k
⇒ k = 3

3. 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. Let A = (0, –1) = , B = (2, 1) =  and
C = (0, 3) =
Area of △ABC =
⇒ Area of △ABC
[0 (1 − 3) + 2 {3 − (−1)} + 0 (−1 − 1)] =
= 4 sq. units

P, Q and R are the mid–points of sides AB, AC and BC respectively.
Applying Section Formula to find the vertices of P, Q and R, we get

Applying same formula, Area of △PQR =   [1 (1 − 2) + 0 (2 − 0) + 1 (0 − 1)] =
= 1 sq. units (numerically)
Now,

4. Find the area of the quadrilateral whose vertices taken in order are (–4, –2), (–3, –5), (3, –2) and (2, 3).
= Area of Triangle ABD +
Area of Triangle BCD … (1)

Using formula to find area of triangle:
Area of  ABD

=  [−4 (−5 − 3) – 3 {3 − (−2)} + 2 {−2 − (−5)}]
(32 – 15 + 6)
(23) = 11.5 sq units … (2)
Again using formula to find area of triangle:
Area of △BCD =
[−3 (−2 − 3) + 3 {3 − (−5)} + 2 {−5 − (−2)}]
(15 + 24 − 6)
(33) = 16.5 sq units … (3)
Putting (2) and (3) in (1), we get
Area of Quadrilateral ABCD = 11.5 + 16.5 = 28 sq units

5. We know that 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. We have △ABC whose vertices are given.
We need to show that ar(△ABD) = ar(△ACD).

Let coordinates of point D are (x, y)
Using section formula to find coordinates of D, we get

Therefore, coordinates of point D are (4, 0)
Using formula to find area of triangle:
Area of △ABD =
[4 (−2 − 0) + 3 {0 − (−6)} + 4 {−6 − (−2)}]
=  (−8 + 18 −16)
=  (−6) = −3 sq units
Area cannot be in negative.
Therefore, we just consider its numerical value.
Therefore, area of △ABD = 3 sq units … (1)
Again using formula to find area of triangle:
Area of △ACD =
[4 (2 − 0) + 5 {0 − (−6)} + 4 {−6 −2 )}]
(8 + 30 − 32) = ½ (6) = 3 sq units … (2)
From (1) and (2), we get ar(△ABD) = ar(△ACD)
Hence Proved.