Question

Find the co-ordinate of the points on the line
$x+5y=13$ at a distance of
$2$ units from the line
$12x-5y+26=0$.

Answer 1

Let the point be $(p,q)$ so that
$p+5q=13$
Its distance from the line $12x-5y+26=0$ is
$\frac{12p-5q+26}{\surd (144+25)}=\pm 2$
Or $12p-5q+26=\pm 26$, given
$+$ive sign $12p-5q=0$
$-$ive sign $12p-5q=-52$
Solving $\left(1\right)$ and $\left(2\right)$, we get the point $(p,q)$ as $(1,12/5)$.
Solving $\left(1\right)$ and $\left(3\right)$, we get the point $(p,q)$ as $(-3,16/5)$.
Alternative method:
Any point on the line $x+5y=13$ be taken as $(13-5t,t)$, choosing $y=t$
Its distance from $12x-5y+26=0$ is $2$
$\therefore \frac{12(13-5t)-5t+26}{\sqrt{144+25}}=\pm 2$
Or $14\times 13-13\left(5t\right)=\pm 2\left(13\right)$
Or $14-5t=\pm 2$ $\therefore 5t=16$ or $12$
Putting for $t$ in $\left(1\right)$, the points are
$(1,\frac{12}{5})$ or $(-3\frac{16}{5})$