Question

Find the coordinates of those points on the line $3x+2y=5$ which are equidistant from the lines $4x+3y=7$ and $2y-5=0$.

Answer 1

Any point which is equidistant from the given lines will lie on the bisector of the lines whose equations are
$\frac{4x+3y-7}{\sqrt{16+9}}=\pm \frac{2y-5}{\sqrt{4+10}}$
or $8x-4y+11=0$ and $8x+16y-39=0$. If the coordinates of the required point be $(\alpha ,\beta )$ then it will liie on the bisectos
$\therefore 8\alpha -4\beta +11=\mathrm{0...}\left(1\right)$
and $\therefore 8\alpha -4\beta +11=\mathrm{0...}\left(1\right)....\left(2\right)$
Since it lies on the line $3x+2y=5$
$\therefore 3\alpha -2\beta =\mathrm{5...}\left(3\right)$
Solving $\left(1\right)$ and $\left(3\right)$ and solving $\left(2\right)$ and $\left(3\right)$ we get the coordinates of the required points as
$(\frac{-1}{14},\frac{73}{28})$ and $(\frac{1}{16},\frac{77}{32})$.