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The vertices of $\triangle ABC$ are $(-2, 1), (5, 4)$ and $(2, -3)$ respectively. Find the area of the triangle and the length of the altitude through $A$.
Given:
The vertices of $\triangle ABC$ are $(-2, 1), (5, 4)$ and $(2, -3)$ respectively.
To do:
We have to find the area of the triangle and the length of the altitude through $A$.
Solution:
Let $AD$ be the length of the altitude through $A$.
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 \( ABC =\frac{1}{2}[-2(4+3)+5(-3-1)+2(1-4)] \)
\( =\frac{1}{2}[-2(7)+5(-4)+2(-3)] \)
\( =\frac{1}{2}[-14-20-6] \)
\( =\frac{1}{2} \times(-40) \)
\( =-20=20 \) sq. units
Length of \( \mathrm{BC}=\sqrt{(2-5)^{2}+(-3-4)^{2}} \)
\( =\sqrt{(-3)^{2}+(-7)^{2}} \)
\( =\sqrt{9+49} \)
\( =\sqrt{58} \) units
Area of \( \Delta \mathrm{ABC}=\frac{1}{2} \) Base \( \times \) Altitude
\( \Rightarrow 20=\frac{1}{2} \sqrt{58} \times AD \)
\( \Rightarrow AD=\frac{20 \times 2}{\sqrt{58}} \)
\( =\frac{40}{\sqrt{58}} \)
The area of the triangle is \( 20 \) sq. units and the length of the altitude through \( A \) is \( \frac{40}{\sqrt{58}} \) units.
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