Question

Let the base of a triangle lie along the line $x=a$ and be of length $a$. The area of this triangle is $a^2$, if the vertex lies on the line

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

Answer 1

The correct option is B $x=2a$

Base of triangle lies along $x=a$

length of base$=2a$

area of triangle is $a^2$

area$=\frac{1}{2}\times$base$\times$height$=a^2$

$=\frac{1}{2}\times 2a\times$ height$=a^2$

$=$ height$=a$

So, third vertex of $\Delta$ is such that the perpendicular on $x=a$ is a.

$\therefore 3^{rd}$ vertex of $\Delta$ lies on $x=0$ or $x=2a$.