Find the area of a triangle whose vertices are$(1, -1), (-4, 6)$ and $(-3, -5)$

Given:

Vertices of a triangle are $(1, -1), (-4, 6)$ and $(-3, -5)$.

To do:

We have to find the area of the given triangle.

Solution:

Let $A(1, -1), B(-4, 6)$ and $C(-3, -5)$ 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 \( ABC=\frac{1}{2}[1(6+5)+(-4)(-5+1)+(-3)(-1-6)] \)

\( =\frac{1}{2}[1(11)+(-4)(-4)+(-3)(-7)] \)

\( =\frac{1}{2}[11+16+21] \)

\( =\frac{1}{2} \times 48 \)

\( =24 \) sq. units.

The area of the given triangle is $24$ sq. units. 

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Updated on: 10-Oct-2022

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