GSEB Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3

Gujarat Board GSEB Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3

Question 1.
Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
Solution:
(i) Let the given point be A(2, 3), B(-1, 0) and C(2, -4).
Here, we have
x₂ = 2 ,y₁= 3
x₂ = -1, y₂ = 0
and x₃ = 2, y₃ = -4
Now, area of ΔABC
= \(\frac {1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\) [2[(0) – (-4)1 +(-1)(-4 -3) + 2(3 – 0)1
= \(\frac {1}{2}\)(8 + 7 + 6) = \(\frac {1}{2}\) x 21
= \(\frac {21}{2}\) sq. units

(ii) Let the given points be A(-5, -1), B(3, -5) and C(5, 2).
Here, we have
x₁ = -5, y₁ = -1
x₂ = 3, y₂ = -5
and x₃ = 5, y₃ = 2
Now, area of ΔABC
= \(\frac {1}{2}\)[x₁(y₁ – y₁) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\)(-5) [(-5-2)] +(3) [2-(-1)] + (5)[(-1) – (-5)]
= \(\frac {1}{2}\) [35 + 9 + 20]
= 32 sq.units

Question 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)
Solution:
(i) Let the given points be A(7, -2), B(5, 1) and C(3, k).
Here, we have,
x₁ = 7 y₁ = -2
x₂ = 5, y₂ = 1
and x₃ = 3, y₃ = k
Now, area of ΔABC
= \(\frac {1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\) 7[(1 -k)] + 5[k -(-2)] + 31(-2 – 1)]
= \(\frac {1}{2}\) [7 – 7k + 5k + 10 – 9]
= \(\frac {1}{2}\) [8 – 2k] = 4 – k
If the points are collinear, then area of the triangle = 0
⇒ 4 – k = 0
⇒ k = 4.

(ii) Let the given point be A(8, 1), B(k, -4) and C(2, -5).
Here, we have
x₁ = 8, y₁ = 1
x₂ = k, y₂ = -4
and x₃ = 2, y₃ = -5
Now, area of ΔABC
= \(\frac {1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\) [8 – 6k + 10]
= [18 – 6k] = 9 – 3k
If the points are collinear, then area of the triangle = 0

Question 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 the area of the triangle formed to area of the given triangle.
Solution:
Let A(0, -1), B(2, 1) and C(0, 3) be the vertices of ABC. Let D, E, F be the mid-points of sides BC, CA and AB respectively.
Then, the coordinates of D, E and F are (1, 2), (0, 1) and (1, 0) respectively.

Now, Area of ΔABC
= \(\frac {1}{2}\)[x₁(y₁ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\) [0(1 – 3) +2[3 -(-1)] + 0(-1-1)]
= \(\frac {1}{2}\) [0 + 8 + 0] = 4 sq. units

Area of ΔDEF
= \(\frac {1}{2}\) [x₁(y₂ -y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Area of ΔDEF
= [1(1 – 0) + 0(0 – 2) + [(2 – 1)]
= Area of ADEF = \(\frac {1}{2}\) [1 + 1]
= 1 sq. unit
∴ Area of DEF : Area of ABC = 1 : 4.

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, 5), (3, -2) and (2, 3). [National Olympiad]
Solution:
Let the vertices of quadrilateral be A(-4, -2), B(-3, -5), C(3, -2) and D(2, 3).
Area of ΔABC
= \(\frac {1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Here, we have
x₁ = -4, y₁ = -2
x₂= -3, y₂ = -5
x₃ = 3, y₃ = -2
Now, area of ΔABC
= \(\frac {1}{2}\) [(-4(-5 + 2) + (-3)(-2 + 2) + 3(-2 + 5)]
= \(\frac {1}{2}\) [(-4 x -3) + (-3 x 0) + (3 x 3)]
= \(\frac {1}{2}\) [12 + 0 + 9] = \(\frac {1}{2}\) x 21
= \(\frac {21}{2}\) sq. units,
We know that, area of ΔACD
= [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Here, we have
x₁ = -4, y₂ = -2
x₂ = 3, y₂ = -2
and x₃ = 2, y₃= 3

Now, area of ΔACD
= \(\frac {1}{2}\) [(-4 x -2 -3) + 3 (3 + 2) + 2 (-2 + 2)]
= \(\frac {1}{2}\) [(-4 x -5) + (3 x 5) + (2 x 0)]
= \(\frac {1}{2}\) [20 + 15 + 0]
= \(\frac {1}{2}\) x 35 = \(\frac {35}{2}\)
Hence, the area of quadrilateral ABCD
= ar(ΔABC) + ar(ΔACD)
= \(\frac {21}{2}\) + \(\frac {35}{2}\) = \(\frac {1}{2}\)(21 + 35)
= \(\frac {1}{2}\) x 56 = 28 sq. units.

Question 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 ΔABC whose vertices are A(4, -6), B(3, -2) and C(5, 2).
Solution:
According to the question, AD is the median of ΔABC, therefore D is the mid-point of BC.
∴ Coordinates of D are \(\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)\), i.e., (4, 0).
Here, for ΔADC,
x₁ = 4, y₁ = 0,
x₂= 4, y₂ = 0,
x₃ = 5, y₃ = 2

Area of ΔABD
= [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\)[4(0 – 2) + 4 (2 + 6) + 5(-6 – 0)]
= \(\frac {1}{2}\)[(-8 + 32 – 30)] = \(\frac {1}{2}\) [-6]
= -3 = 3 sq. units
[ ∴ Area of triangle is positive and Area of ΔABD]
= \(\frac {1}{2}\) [4(-2 -0) + 3(0 + 6) + 4(-6 + 2)]

For ΔABD,
[Let, x₁ = 4, y₁ = -6, x₂ = 3, y₂ = -2, x₃ = 4, y₃ = 0]
= [-8 + 18 – 16] = (-6) = -3
= 3 sq. units [∴ Area of triangle is positive.]
Area of ADC = Area of AED
Hence, the median of the triangle divides it into two triangles of equal areas.