Question

The coordinates of $A$ for which area of triangle, whose vertices are $A(a,2a),B(-2,6)$ and $C(3,1)$ is $10$ square units, are:

• A. $(0,3)$
• B. $(5,8)$
• C. $(3,\frac{8}{3})$
• D. None of these

Answer 1

The correct option is D None of these

Given: Vertices of the triangle are $A(a,2a),B(-2,6),C(3,1)$
The area of the triangle is $10$ square units.

$\Rightarrow \frac{1}{2}|a(6-1)+(-2)(1-2a)+3(2a-6)|=10$

$\Rightarrow 5a-2+4a+6a-18=20$

$\Rightarrow 15a=40$

$\Rightarrow a=\frac{40}{15}=\frac{8}{3}$

$\therefore$ The coordinates of vertex $A$ are $(\frac{8}{3},\frac{16}{3})$