Question

Find the coordinates of the point at unit distance from the lines $3x-4y+11=0$ and $5x-12y+7=0$.

Answer 1

Let the given lines be $l_1$ and $l_2$ which have two bisectors of angle as $B_1$ and $B_2$. Any point on the bisector is equidistant from the two lines
Thus we will have four points like $P_1,P_2,P_3$ and $P_4$
Let $P(x,y)$ which is equidistant
$\frac{3x-4y+11}{\sqrt{3^2+4^2}}=1;\frac{5x-12y+7}{\sqrt{5^2+{12}^2}}=1$
or $3x-4y+11=\pm 5$, $5x-12y+7=\pm 13$
or $3x-4y+16=\mathrm{0.....}\left(1\right)$
$3x-4y+6=\mathrm{0.......}\left(2\right)$
and $5x-12y+20=\mathrm{0....}\left(3\right)$
$5x-12y-6=\mathrm{0......}\left(4\right)$
Solving (1) with (3) and (4), and then (2) with (3) and (4) we get the points $(-7,-5/4),(-27/2,-49/8),(1/2,15/8),(-6,-3)$