Question

The ends of the base of an isosceles triangle are at $\left(2a,0\right)$ and $\left(0,a\right)$. The equation of one side is $x=2a$.The equation of the other side is

• A. $x+2y-a=0$
• B. $x+2y=2a$
• C. $3x+4y-4a=0$
• D. $3x-4y+4a=0$

Answer 1

The correct option is D

$3x-4y+4a=0$

Explanation for the correct option:

Assume $A$, $B,C$are the given vertices of the triangle

$AB=AC$ (as the triangle is isosceles)

Using distance formula

$$\begin{gathered} \Rightarrow \sqrt{{\left(2a-0\right)}^2+{\left(h-a\right)}^2}=\sqrt{{\left(2a-2a\right)}^2+{\left(h-0\right)}^2} \\ \Rightarrow 4a^2+h^2+a^2-2ah=h^2 \\ \Rightarrow 5a^2=2ah \\ \Rightarrow 5a=2h \\ \Rightarrow h=\frac{5a}{2} \end{gathered}$$

So, the equation of the line $AB$is:

$$\begin{gathered} \frac{y-a}{{\displaystyle \frac{5a}{2}}-a}=\frac{x-0}{2a-0} \\ \Rightarrow 4ay-4a^2=3ax \\ \Rightarrow 4y-4a=3x \\ \Rightarrow 3x-4y+4a=0 \end{gathered}$$

Hence, option (D) is the correct answer