Question

If the area of a triangle is $68$ sq. units and the vertices are $(6,7),(-4,1)$ and $(a,-9)$ then the value of $a$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Area of a triangle with vertices $(x_1,y_1)$ ; $(x_2,y_2)$ and $(x_3,y_3)$ is:

$A=\left|\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\right|$
Hence, substituting the points $(x_1,y_1)=(6,7)$ ; $(x_2,y_2)=(-4,1)$ and $(x_3,y_3)=(a,-9)$ in the formula for area, we get:

Area of triangle $=\left|\frac{\left(6\right)(1+9)+(-4)(-9-7)+a(7-1)}{2}\right|=68$
$\left|\frac{60+64+6a}{2}\right|=68$
$\frac{124+6a}{2}=68$
$124+6a=136$
$6a=12$
$⟹a=2$