Question

Find the area of a triangle ABC whose vertices are A(-2 ,2) B(5 ,2) and whose centroid is (1 , 3)

• A. 21/2
• B. 23/2
• C. 19/2
• D. 25/2

Answer 1

The correct option is A 21/2

Centroid of a triangle with vertices ${(x}_1,y_1);{(x}_2,y_2)$ and ${(x}_3,y_3)$ is calculated by the formula $(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$
Let the third vertex of the triangle be $C(x,y)$
So, centroid $=(\frac{-2+5+x}{3},\frac{2+2+y}{3})=(1,3)$
$=(\frac{3+x}{3},\frac{4+y}{3})=(1,3)$
$=>3+x=3;4+y=9$
$x=0;y=5$
Hence the third vertex of the triangle is $(0,5)$
Area of a triangle with vertices $(x_1,y_1)$ ; $(x_2,y_2)$ and $(x_3,y_3)$ is $\left|\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\right|$
Hence, substituting the points $(x_1,y_1)=(-2,2)$ ; $(x_2,y_2)=(5,2)$ and $(x_3,y_3)=(0,5)$ in the area formula, we get
Area of triangle ABC $=\left|\frac{(-2)(2-5)+\left(5\right)(5-2)+0(2-2)}{2}\right|=\left|\frac{6+15}{2}\right|=\frac{21}{2}squnits$