Question

Find the value of k so that the area of the triangle with vertices A(k+1,1),B(4,-3) and C(7, -k) is 6 square units.

Answer 1

Let $A(x_1,y_1)=A(k+1,1),B(x_2,y_2)=B(4,-3)andC(x_3,y_3)=C(7,-k)$.

Now

$Area\left(\Delta ABC\right)=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]\Rightarrow 6=\frac{1}{2}\left[\right(k+1\left)\right(-3+k)+4(-k–1)+7(1+3\left)\right]\Rightarrow 12=[k^2–2k–3–4k–4+28]\Rightarrow 12=k^2–6k–21\Rightarrow k^2–6k–9=0\Rightarrow (k-3{)}^2=0\Rightarrow k-3=0\Rightarrow k=3$