Question

For the straight lines 4x+3y-5 = 0 and 12x-5y+6 = 0, find the equation of the bisector which contains (1,3)

• A. 8x+9y = 95
• B. 8x-64y = 95
• C. 8x-64y+95 = 0
• D. 8x+9y+95 = 0

Answer 1

The correct option is C

8x-64y+95 = 0

Equation of the bisector of $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ which contains the point $(\alpha ,\beta$ is given by .

$\frac{a1x+b1y+c1}{\sqrt{a_1^2+b_1^2}}=\frac{a2x+b2y+c2}{\sqrt{a_2^2+b_2^2}}$ if $a_1\alpha +b_1\beta +c_1$ and $a_2\alpha +b_2\beta +c_2$ have same sign , otherwise by

$\frac{a1x+b1y+c1}{\sqrt{a_1^2+b_1^2}}=\frac{a2x+b2y+c2}{\sqrt{a_2^2+b_2^2}}$

Our point is (1,3) substituting it in the given lines give 4*1+3*3-5 and 121-8*3+6 , or 8 and 3 , which are of same sign.

$\Rightarrow$ +ve sign gives the bisector.

$\Rightarrow$ $\frac{4x+3y-5}{5}=\frac{12x-5y+6}{13}$

52x + 39y - 65 = 60x - 25y + 30

8x - 64y + 35 = 0