TS 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3

Students can practice TS Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.3

Question 1.
Find the area of the triangle whose vertices are

i) (2, 3), (-1, 0), (2,-4)
Solution:
Given : A(2, 3), B(-1, 0), C(2, -4) are the vertices of a ΔABC
Area of the ΔABC
= \(\frac{1}{2}\)|(y₂ – y₃) + x₂ (y₃ – y₁) + x₃(y₁ – y₂)|
= \(\frac{1}{2}\) |8 + 7 + 6|
= \(\frac{21}{2}\) = 10\(\frac{1}{2}\) sq. units.

ii) (- 5, – 1), (3, -5), (5, 2) (A.P. Mar. 15)
Solution:
Given A(-5, -1), B(3, -5), C(5, 2) are the ver-tices of a ΔABC
Area of the Δ ABC
= \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) | 135 + 9 + 20|
= \(\frac{64}{2}\) = 32 sq.units.

iii) (0, 0), (3, 0) and (0, 2)
Solution:
Given 0(0, 0), A(3, 0), B(0, 2) are the vertices of a ΔABC
Area of the ΔAOB
= \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) |0(0 – 2) + 3(2 – 0) + 0(0 – 0)|
= \(\frac{1}{2}\) |6| = 3 sq. units. (or)

ΔAOB = \(\frac{1}{2}\) × 3 × 2 = 3 sq. units.

Question 2.
Find the value of ‘K’ for which the points are collinear. (A.P. June ’15)

i) (7, -2), (5, 1), (3, K)
Solution:
Given : A(7, -2), B(5, 1), C(3, K) are collinear
∴ Area of ΔABC = 0
But area of triangle.
⇒ \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
⇒ \(\frac{1}{2}\) | 7 – 7K + 5K + 10 – 9| = 0
⇒ |-2K + 8| = 0
⇒ – 2K = – 8
⇒ K = 4

ii) (8, 1), (K, -4), (2, -5)
Solution:
Given A(8, 1), B(k, -4), C(2, -5) are collinear
∴ Area of ΔABC = 0
⇒ \(\frac{1}{2}\) |8(- 4 + 5) + K(-5 – 1) + 2 (1 + 4) 1= 0
⇒ | 8 – 6K + 10| = 0
⇒ 18 – 6K = 0
⇒ 6K = 18
⇒ K = \(\frac{18}{6}\)
∴ K = 3

iii) (K, K) (2, 3) and (4, -1)
Solution:
A(K, K), B(2, 3) & C(4, -1) are collinear
∴ Area of ∆ABC = O
⇒ \(\frac{1}{2}\) | K(3 + 1) + 2 (-1 – K) + 4(K – 3)| = 0
⇒ | 4K – 2 – 2k + 4K – 12| = 0
⇒ |6K – 14| = 0
⇒ 6K = 14
⇒ K = \(\frac{14}{6}\)
= \(\frac{7}{3}\)
⇒ K = \(\frac{7}{3}\)

Question 3.
Find the area of the triangle formed by joining the midpoints 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. (A.P. Mar.16)
Solution:

Given : A(0, -1), B(2, 1), C(0, 3) are the vertices of ∆ABC. Let D, E & F be the midpoints of the sides \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\) & \(\overline{\mathrm{AC}}\)
midpoint (x, y) = \(\left[\frac{\mathrm{x}_1+\mathrm{x}_2}{2}, \frac{\mathrm{y}_1+\mathrm{y}_2}{2}\right]\)
∴ D = \(\left(\frac{0+2}{2}, \frac{-1+1}{2}\right)\) = (1, 0)
E = \(\left(\frac{2+0}{2}, \frac{1+3}{2}\right)\) = (1, 2)
F = \(\left(\frac{0+0}{2}, \frac{-1+3}{2}\right)\) = (0, 1)
Area of ∆DEF
= \(\frac{1}{2}\) |1(2 – 1) + 1(1 – 0) + 0(0 – 2)|
= \(\frac{1}{2}\) |1 + 1| = \(\frac{2}{2}\) = 1 sq. units.
Ratio of areas = ∆ABC : ∆DEF = 4 : 1
∆ADF \(\cong\) ∆BED \(\cong\) ∆DEF \(\cong\) ∆CEF
∴ ∆ABC : ∆DEF = 4 : 1

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3)
Solution:

Given : A(-4, -2), B(-3, -5), C(3, -2) & D(2, 3) are the vertices of the quadrilateral ABCD.
Area of
ABCD = ∆ABC + ∆ACD
Area of triangle
= \(\frac{1}{2}\) |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
Area of ∆ABC
= \(\frac{1}{2}\) |(-4)(-5 + 2)(-3)(-2 + 2) + 3(-2 + 5) |
= \(\frac{1}{2}\)|12 – 0 + 9| = \(\frac{21}{2}\) sq.units.
Area of ∆ACD
= \(\frac{1}{2}\)|-4(-2 – 3) + 3(3 + 2) + 2(-2 + 2) |
= \(\frac{1}{2}\)|20 + 15 + 0 | = \(\frac{35}{2}\) sq. units.
∴ Area of ABCD = \(\frac{21}{2}\) + \(\frac{35}{2}\) = \(\frac{21+35}{2}\)
= \(\frac{56}{2}\) = 28 sq.units.

Question 5.
Find the area of the triangle formed by the points (8, -5), (-2, -7) & (5, 1) by using Herons formula.
Solution:
Given : A(8, -5), B(-2, -7) & C(5, 1) are the vertices of ∆ABC
Distance formula
= \(\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)
AB = \(\sqrt{100+4}\) = \(\sqrt{104}\) = 10.19 = a
BC = \(\sqrt{49+64}\) = \(\sqrt{113}\) = 10.63 = b