Question

Find the points on the line $x+y=7$ at a distance of $5$ units from $P(1,3)$.

Answer 1

Point is $(1,3)$

Distance is $5$

$x+y=7tan\theta =-11\sin \theta =-\frac{1}{\sqrt{2}}\cos \theta =\frac{1}{\sqrt{2}}$

The required point is $(x_1\pm r\sin \theta ,y_1\pm r\cos \theta )(1\pm 5\frac{-1}{\sqrt{2}},3\pm 5\frac{1}{\sqrt{2}})(1\mp \frac{5}{\sqrt{2}},3\pm \frac{5}{\sqrt{2}})$