Gujarat Board Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3
Gujarat Board Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3
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
= 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= 1/2 [2[(0) – (-4)1 +(-1)(-4 -3) + 2(3 – 0)1
= 1/2(8 + 7 + 6) = 1/2 x 21
= 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
= 1/2[x1(y1 – y1) + x2(y3 – y1) + x3(y1 – y2)]
= 1/2(-5) [(-5-2)] +(3) [2-(-1)] + (5)[(-1) – (-5)]
= 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
= 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= 1/2 7[(1 -k)] + 5[k -(-2)] + 31(-2 – 1)]
= 1/2 [7 – 7k + 5k + 10 – 9]
= 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
= 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= 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
= 1/2 [x1(y1 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= 1/2 [0(1 – 3) +2[3 -(-1)] + 0(-1-1)]
= 1/2 [0 + 8 + 0] = 4 sq. units
Area of ΔDEF
= 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 = 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
= 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
= 1/2 [(-4(-5 + 2) + (-3)(-2 + 2) + 3(-2 + 5)]
= 1/2 [(-4 x -3) + (-3 x 0) + (3 x 3)]
= 1/2 [12 + 0 + 9] = 1/2 x 21
= 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
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.
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.
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