Question

If $y=x$ is the angle bisector of the lines $3x-4y+a=0$, $4x-3y-a=0$ containing the points $(1,0)$, then $a$ can be-

• A. $0$
• B. $2$
• C. $-2$
• D. $-4$

Answer 1

Answer: (B), (C)

$-2$
$2$

Given

$x-y=0$ is angle Bisector of the given lines

$L_1:3x-4y+a=0$__ (i)

$L_2:4x-3y-a=0$ __ (ii)

$x-y=0$

$(1,0)=1>0$

$L_1(1,0)>0$

$(3-yx+0+a)>0$

$\Rightarrow a>-3$ __ (iii)

$L_2(1,0)>0$

$y-a>0\Rightarrow 4>a$

$\therefore a<4$ __ (iv)

from (iii) & (iv)

$aϵ(-3,4)$

$\therefore a=-2$

$a=2$