Question

In an isosceles triangle, the ends of the base are $(2a,0)$ and $(0,a)$ and one side is parallel to $y$-axis. The third vertex is

• A. $\left(2a,\frac{5a}{2}\right)$
• B. $\left(\frac{5a}{4},a\right)$
• C. $(2a,a)$
• D. $(4a,a)$

Answer 1

Answer: (A)

$\left(2a,\frac{5a}{2}\right)$

So, mid point of $(0,a)$ & $(2a,0)$ is $(a,\frac{a}{2})$

$\Rightarrow$ perpendicular bisector of AB passes through $(a,\frac{a}{2})$

And having slope $=2$

So, equation of line is

$y-\frac{a}{2}=2(x-a)$

$2y-a=4x-4a$

$2y-4x=-3a$ ---(2)

$LineL:2y-4x=-3a$ intersect with this side $x=2a$

$\Rightarrow$ gives $x=2a$ and $y=\frac{5a}{2}$