For what value of k (k > 0) is the area of the triangle with vertices (−2, 5), (k, −4) and
Question:

For what value of k (k > 0) is the area of the triangle with vertices (−2, 5), (k, −4) and
(2k + 1, 10) equal to 53 square units?

Solution:

Let $A\left(x_{1}=-2, y_{1}=5\right), B\left(x_{2}=k, y_{2}=-4\right)$ and $C\left(x_{3}=2 k+1, y_{3}=10\right)$ be the vertices of

the triangle. So

Area $(\Delta A B C)=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$\Rightarrow 53=\frac{1}{2}[(-2)(-4-10)+k(10-5)+(2 k+1)(5+4)]$

$\Rightarrow 53=\frac{1}{2}[28+5 k+9(2 k+1)]$

$\Rightarrow 28+5 k+18 k+9=106$

$\Rightarrow 37+23 k=106$

$\Rightarrow 23 k=106-37=69$

$\Rightarrow k=\frac{69}{23}=3$

Hence, k = 3.

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