Question

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

• A. $21x–77y+100=0$
• B. $99x–27y+81=0$
• C. $99x+27y+30=0$
• D. $21x–77y–100=0$

Answer 1

The correct option is B

$99x–27y+81=0$

Step 2. Find the discriminant of the given equations:

Given: $3x–4y+7=0$ ..…(i)

$12x+5y–2=0$

$\Rightarrow$ $-12x–5y+2=0$ ..…(ii)

The discriminant of the given equations

$a_1{a}_2+b_1{b}_2=-36+20$

$=-16<0$

Step 2. Find the acute angle bisector:

acute angle bisector will be,

$\frac{(a_{1}x+b_{1}y+c_1)}{\sqrt{({a_1}^2+{b_1}^2)}}=\frac{(a_{2}x+b_{2}y+c_2)}{\sqrt{({a_2}^2+{b_2}^2)}}$

$\Rightarrow$ $\frac{(3x–4y+7)}{\sqrt{(3^2+4^2})}=\frac{(-12x–5y+2)}{\sqrt{({12}^2+5^2)}}$

$\Rightarrow$ $\frac{(3x–4y+7)}{\sqrt{25}}=\frac{(-12x–5y+2)}{\sqrt{169}}$

$\Rightarrow$ $\frac{(3x–4y+7)}{5}=\frac{(-12x–5y+2)}{13}$

$\Rightarrow$ $39x–52y+91=-60x–25y+10$

$\Rightarrow$$99x–27y+81=0$

Hence, Option ‘B’ is Correct.