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

The vertices of the given $△$ ABC are A(1,-1), B(-4,2k) and C(-k,-5)

$\therefore \text{Area of}△ABC=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

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

$=\frac{1}{2}[2k+5+16+k+2k^2]$

=$\frac{1}{2}[2k^2+3k+21]$

Area of $△ABC$ =24 sq.units (Given)

$\therefore \frac{1}{2}[2k^2+3k+21]=24$

$\Rightarrow 2k^2+3k+21=48$

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

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

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

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

k = 3 or -$\frac{9}{2}$

Hence, k = 3 or -$\frac{9}{2}$