Question

Find the area (in square units) of the triangle whose vertices are $(a,b+c),(a,b-c)$ and $(-a,c).$

• A. $2ac$
• B. $2bc$
• C. $b(a+c)$
• D. $c(a-h)$

Answer 1

The correct option is A $2ac$

Area of triangle having vertices $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ is given by
Area $=\frac{1}{2}\times |\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]|$

Given the vertices of triangle,

$A(a,b+c),B(a,b-c),C(-a,c)$

Therefore, area is given by

$=\frac{1}{2}|[a[b-c-c]+a[c-b-c]+(-a\left)[b+c-b+c]\right]|=\frac{1}{2}|\left[a\right(b-2c)+a(-b)-a(2c\left)\right]|=\frac{1}{2}|[ab-2ac-ab-2ac]|=\left|\frac{-4ac}{2}\right|=2ac$