Find the area of the triangle formed by the points.
(i) (0, 0), (3, 0) and (0, 2)
Solution :
Let A(0, 0) , B(3, 0) and C(0, 2) be the vertices of the triangle taken in order.
x1 = 0 x2 = 3 x3 = 0
y1 = 0 y2 = 0 y3 = 2
Area of the triangle triangle ABC :
= (1/2)[(0 + 6 + 0) – (0 + 0 + 0)]
= (1/2)[6 - 0]
= (1/2)[6]
= 3 square units
(ii) (5, 2) (3, -5) and (-5, -1)
Solution :
Let A (-5, -1), B (3, -5) and C (5, 2) be the vertices of the triangle taken in order.
x1 = -5 x2 = 3 x3 = 5
y1 = -1 y2 = -5 y3 = 2
Area of the triangle ABC :
= (1/2)[(25 + 6 - 5) – (-3 - 25 - 10)]
= (1/2)[(31 - 5) –(-38)]
= (1/2)[26 +38]
= (1/2)[64]
= 32 square units
(iii) (-4, -5), (4, 5) and (-1, -6)
Solution :
Let A(-4, -5) , B(-1, -6) and C(4, 5) are the vertices of the triangle taken in order.
x1 = -4 x2 = -1 x3 = 4
y1 = -5 y2 = -6 y3 = 5
Area of the triangle ABC :
= (1/2)[(24 - 5 - 20) – (5 - 24 - 20)]
= (1/2)[(24 - 25) –(5 - 44)]
= (1/2)[(-1) – (-39)]
= (1/2)[-1 + 39]
= (1/2)[38]
= 19 square units
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Dec 21, 24 02:20 AM
Dec 21, 24 02:19 AM
Dec 20, 24 06:23 AM