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For what value of $k(k>0)$ is the area of the triangle with vertices $(-2,5),(k,-4)$ and $(2 k+1,$, 10) equal to 53 square units?
For what value of $k(k>0)$ is the area of the triangle with vertices $(-2,5),(k,-4)$ and $(2 k+1,$, 10) equal to 53 square units?
CBSE | Class 10 | Coordinate Geometry | Exercise 16.3 | Maths | RS Aggarwal
For what value of $k(k>0)$ is the area of the triangle with vertices $(-2,5),(k,-4)$ and $(2 k+1,$, 10) equal to 53 square units?

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

$$
\begin{aligned}
&\text { Area }(\Delta \mathrm{ABC})=\frac{1}{2}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right] \\
&\Rightarrow 53=\frac{1}{2}[(-2)(-4-10)+\mathrm{k}(10-5)+(2 \mathrm{k}+1)(5+4)] \\
&\Rightarrow 53=\frac{1}{2}[28+5 \mathrm{k}+9(2 \mathrm{k}+1)] \\
&\Rightarrow 28+5 \mathrm{k}+18 \mathrm{k}+9=106 \\
&\Rightarrow 37+23 \mathrm{k}=106 \\
&\Rightarrow 23 \mathrm{k}=106-37=69 \\
&\Rightarrow \mathrm{k}=\frac{69}{23}=3
\end{aligned}
$$

Therefore, $\mathrm{k}=3$.

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