Find the points on the line 4x –3y – 10 = 0 which are at a distance of 5 units from the point (1,–2).

The points on the line which are at a distance ‘r’ units from P(x₁, y₁) are
$\mathrm{x}={\mathrm{x}}_1\pm \mathrm{rcos}\mathrm{\theta },\mathrm{y}={\mathrm{y}}_1\pm \mathrm{rsin}\mathrm{\theta }$
$\text{ given st.line }4\mathrm{x}-3\mathrm{y}-10=0 ...\left(1\right)$
$\text{ slope of (1) is }\left(\mathrm{m}\right)=\mathrm{Tan}\mathrm{\theta }=\frac{4}{3}$
$\sin \mathrm{\theta }=\frac{4}{5},\cos \mathrm{\theta }=\frac{3}{5}$
And given st.line is at a distance of 5 units from the point (1, –2)  ,so r=5
$\begin{array}{l}\mathrm{Now} \mathrm{x}=1\pm 5\left(\frac{3}{5}\right), \mathrm{y}=-2\pm 5\frac{4}{5}\\ x=1+3\mathrm{\&}1-3 , y=-2+4\mathrm{\&}-2-4\\ x=4 \mathrm{\&}-2 ,y=2 \mathrm{\&}-6\end{array}$
$\therefore \text{ the points on the line is }(4,2)\text{ and }(-2,-6)$