Question

The base of a triangle lies along $x=a$ and is of length $a$. If the area of the triangle is $a^2$, then the locus of its vertex is

• A. $(x+a)(x-3a)=0$
• B. $(x-a)(x+3a)=0$
• C. $(x+a)(x+3a)=0$
• D. $(x+2a)(x-a)=0$

Answer 1

The correct option is A $(x+a)(x-3a)=0$

Suppose, the coordinates of the vertex are $(h,k)$.
The area of the triangle is $\frac{1}{2}\times a|h-a|=a^2$.
$\Rightarrow h=-a$ or $h=3a$.
Hence, the combined equation of the pair of lines is $(h+a)(h-3a)=0$
$(x+a)(x-3a)=0$ is the required locus.