Question

Find the points on the x-axis, whose distance from the line $\frac{x}{3}+\frac{y}{4}=1$ are $4$ units.

Answer 1

$Pointonx-axisis(x,0)Dis\tan cefromline=\frac{\left|\frac{x1}{3}+\frac{y1}{4}-1\right|}{\sqrt{\frac{1}{3^3}+\frac{1}{4^2}}}4=\frac{\left|\frac{x}{3}+0-1\right|}{\sqrt{\frac{1}{9}+\frac{1}{6}}}4=\frac{\left|\frac{x}{3}-1\right|}{\sqrt{\frac{25}{9\times 16}}}4\times \sqrt{\frac{25}{9\times 16}}=\frac{x}{3}-14\times \frac{5}{3\times 4}=\frac{x}{3}-1\frac{5}{3}+1=\frac{x}{3}=\frac{8}{3},\therefore x=8Pointis(8,0)$