The equation of the bisector of the acute angle between the lines 3 x-4 y+7=0 and 12 x+5 y-2=0 is

Question

The equation of the bisector of the acute angle between the lines 3 x-4 y+7=0 and 12 x+5 y-2=0 is (A) 11 x-3 y+9=0 (B) 3 x+11 y-13=0 (C) 3 x+11 y-3=0 (D) 11 x-3 y+2=0

Tardigrade Admin · Accepted Answer

after making constant terms positive, equation of lines are

3 x - 4 y + 7 = 0.....

(i)

- 12 x - 5 y + 2 = 0.....

(ii)

∵ a_{1} a_{2} + b_{1} b_{2} = - 36 + 20 < 0

∴

equation of acute angle bisector is

\frac{3 x - 4 y + 7}{5} = + \frac{- 12 x - 5 y + 2}{13}

\Rightarrow 11 x - 3 y + 9 = 0

Q. The equation of the bisector of the acute angle between the lines $3 x - 4 y + 7 = 0$ and $12 x + 5 y - 2 = 0$ is

$11 x - 3 y + 9 = 0$

$3 x + 11 y - 13 = 0$

$3 x + 11 y - 3 = 0$

$11 x - 3 y + 2 = 0$

Q. The equation of the bisector of the acute angle between the lines 3x−4y+7=0 and 12x+5y−2=0 is

11x−3y+9=0

3x+11y−13=0

3x+11y−3=0

11x−3y+2=0

after making constant terms positive, equation of lines are 3x−4y+7=0..... (i) −12x−5y+2=0..... (ii) ∵a1​a2​+b1​b2​=−36+20<0 ∴ equation of acute angle bisector is 53x−4y+7​=+13−12x−5y+2​ ⇒11x−3y+9=0

Q. The equation of the bisector of the acute angle between the lines $3 x - 4 y + 7 = 0$ and $12 x + 5 y - 2 = 0$ is

$11 x - 3 y + 9 = 0$

$3 x + 11 y - 13 = 0$

$3 x + 11 y - 3 = 0$

$11 x - 3 y + 2 = 0$