Question

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

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

Answer 1

Answer: (D)

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

Answer 2

Obviously, other line AB will pass through (0, a) and $( 2 a , k )$.

But as we are given $A B = A C$

$\Rightarrow k = \sqrt { 4 a ^ { 2 } + ( k - a ) ^ { 2 } }$⇒ $k = \frac { 5 a } { 2 }$

Hence the required equation is $3 x - 4 y + 4 a = 0$.