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

Given:

Vertices of a triangle are \( (-8,4),(-6,6) \) and \( (-3,9) \).

To do:

We have to find the area of the given triangle.

Solution:

Let $A(-8, 4), B(-6, 6)$ and $C(-3, 9)$ be the vertices of a $\triangle ABC$.

We know that,

Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,

Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Therefore,

Area of triangle \( ABC=\frac{1}{2}[-8(6-9)+(-6)(9-4)+(-3)(4-6)] \)

\( =\frac{1}{2}[-8(-3)+(-6)(5)+(-3)(-2)] \)

\( =\frac{1}{2}[24-30+6] \)

\( =\frac{1}{2} \times 0 \)

\( =0 \) sq. units.

The area of the given triangle is $0$ sq. units. 

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Updated on: 10-Oct-2022

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