Question

Write the equation of the internal bisector of the angle between the lines
$2x-3y-5=0,6x-4y+7=0$ which is the supplement of the angle containing the point $(2,1)$.

Answer 1

Substitute the coordinates of the point $(2,-1)$ in the left hand side of the given equations, the results are $2$ and $23$ i.e., both +ive. Hence, the bisector of the angle in which $(2,-1)$ lies is obtained by taking +ive out of $\pm$. Therefore the equation of bisector of the supplement of that angle is obtained by taking -ive sign. Hence the required equation is
$\frac{2x-3y-5}{\sqrt{4+9}}=-\frac{6x-4y+7}{\sqrt{36+16}}$
or $10x-10y-3=0$