Question

Find the values of $k$ so that the area of the triangle with vertices $(1,-1),(-4,2k)$ and $(-k,-5)$ is $24$ sq. units.

Answer 1

Given area of triangle whose vertices are $(1,-1),(-4,2k)$ and $(-k,-5)$ is $24$ sq units

We know that the area of triangle $=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

$\therefore 24=\frac{1}{2}\left[1\right(2k-(-5))-4(-5-(-1))+(-k\left)\right(-1-2k\left)\right]$

$\Rightarrow 48=[2k+5+(-4\left)\right(-4)+k(1+2k\left)\right]$

$\Rightarrow 48=[2k+5+16+k+2k^2]$

$\Rightarrow 2k^2+3k+5+16-48=0$

$\Rightarrow 2k^2+3k-27=0$

$\Rightarrow 2k^2+9k-6k-27=0$

$\Rightarrow k(2k+9)-3(2k+9)=0$

$\Rightarrow (k-3)(2k+9)=0$

$\Rightarrow k-3=0\text{or}2k+9=0$
$\Rightarrow k=3\text{or}k=\frac{-9}{2}$