If the vertices of a triangle are $(1, -3), (4, p)$ and $(-9, 7)$ and its area is 15 sq. units, find the value(s) of $p$.
Given:
The vertices of a triangle are $(1, -3), (4, p)$ and $(-9, 7)$ and its area is 15 sq. units.
To do:
We have to find the value(s) of $p$.
Solution:
Let $A(1, -3), B(4, p)$ and $C(-9, 7)$ be the vertices of $\triangle ABC$.
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 \( ABC=\frac{1}{2}[1(p-7)+4(7+3)+(-9)(-3-p)] \)
\( 15=\frac{1}{2}[p-7+4(10)+27+9p] \)
\( 15(2)=(10p+20+40) \)
\( 30=10p+60 \)
\( 10p=-60+30 \)
\( p=\frac{-30}{10} \)
\( p=-3 \)
The value of $p$ is $-3$.
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