Question

Question 9
Find the area of the triangle whose vertices are (-8,4), (-6,6) and (-3,9).

Answer 1

Given that, the vertices of triangles are (-8,4), (-6,6) and (-3,9).
$Let(x_1,y_1)\to (-8,4)(x_2,y_2)\to (-6,6)and(x_3,y_3)\to (-3,9)\text{We know that, the area of triangle with vertices}(x_1,y_1),(x_2,y_2)and(x_3,y_3)\Delta =\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=\frac{1}{2}[-8(6-9)-6(9-4)+(-3\left)\right(4-6\left)\right]=\frac{1}{2}\left[\right[-8(-3)-6\left(5\right)-3(-2)]=\frac{1}{2}(24-30+6\left)\right]=\frac{1}{2}(30-30)=\frac{1}{2}\left(0\right)=0\text{Hence, the required area of triangle is 0.}$