Find the area of a triangle whose vertices are$(6, 3), (-3, 5)$ and $(4, -2)$

Given:

Vertices of a triangle are $(6, 3), (-3, 5)$ and $(4, -2)$.

To do:

We have to find the area of the given triangle.

Solution:

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

\( =\frac{1}{2}[6 \times 7+(-3)(-5)+4(-2)] \)

\( =\frac{1}{2}[42+15-8] \)

\( =\frac{1}{2} \times 49 \)

\( =\frac{49}{2} \) sq. units.

The area of the given triangle is $\frac{49}{2}$ sq. units.

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Updated on: 10-Oct-2022

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