Question

find the area of triangle having vertices x( 4, - 3), y( - 6, - 3), z (0, -3) justify your answer

Answer 1

Answer:

0

Step-by-step explanation:

Given vertices are x(4,-3), y(-6,-3) and z(0,-3)

Here, (x₁,y₁) = (4,-3), (x₂,y₂) = (-6,-3) and (x₃,y₃) = (0,-3).

∴ Area of triangle xyz = (1/2)[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

= (1/2)[4(-3+3) + (-6)(-3 + 3) + 0(4+3)]

= (1/2)[0]

= 0.

Hope it helps!

Answer 2

AnSwEr :-


We know in this form to find the area of a triangle follow this formula :-


$x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)$





so, here

x1 = 4

x2 = -6

x3 = 0

y1 = -3

y2 = -3

y3 = -3


following the above formula :-


$4( - 3 - ( - 3)) + - 6( - 3 - ( - 3)) + 0( - 3 - ( - 3))$




$4( - 3 + 3) + - 6( - 3 + 3) + 0( - 3 + 3)$



$(4 \times 0) + ( - 6 \times 0) + (0 \times 0)$


$0 + 0 + 0 = 0$




so, the area is 0


Hope it helps you


$<b > prabhudutt$