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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