Question

find the equation of the bisector of the angle between the lines x + 2y -11=0 , 3x-6y-5=0 which contains the point (1,-3) ​

Answer 1

Answer:

No Solution

Step-by-step explanation:

The equations of bisectors of angle between 2 straight lines ax+by+c=0 and px+qy+r=0 are given by the formula:

$\frac{ax+by+c}{\sqrt{a^{2}+b^{2}} }$=±$\frac{px+qy+r}{\sqrt{p^{2}+q^{2}} }$ ........(1)

Now, we have two equations of two straight lines:

x+2y-11=0 and ......(2)

3x-6y-5=0 ......(3)

So, using equation (1) we get the equations of bisectors of angle between equation(2) and (3):

$\frac{x+2y-11}{\sqrt{1^{2}+2^{2}} }$=±$\frac{3x-6y-5}{\sqrt{3^{2}+(-6)^{2}} }$

⇒$\frac{x+2y-11}{\sqrt{5} }$=±$\frac{3x-6y-5}{\sqrt{45} }$

⇒ 3x+6y-33=±(3x-6y-5)

⇒12y=28 and 6x=38

⇒3y=7 and 3x=19 ....(4)

So, none of the equations (4) is satisfied by (1,-3) point.

Therefore, there is no bisector equation that passes through (1,-3) point.

Hence, no solution.