Question

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

Answer 1

Let the coordinates of the point on x - axis be $(\alpha ,0)$

$\therefore$ Perpendicular distance of the point $(\alpha ,0)$ the line 4x + 3y - 12 = 0 is

$\left|\frac{4\alpha +3\left(0\right)-12}{\sqrt{(4{)}^2+(3{)}^2}}\right|=\left|\frac{4\alpha -12}{\sqrt{16+9}}\right|=\left|\frac{4\alpha -12}{5}\right|$

It is given that $\left|\frac{4\alpha -12}{5}\right|=4$

$\Rightarrow \frac{4\alpha -12}{5}=\pm 4$

Now $\frac{4\alpha -12}{5}=4$ or $\frac{4\alpha -12}{5}=-4$

$\Rightarrow 4\alpha -12=20$ or $4\alpha -12=-20$

$\Rightarrow 4\alpha =32$ or $4\alpha =-8$

$\Rightarrow \alpha =8$ or $\alpha =-2$

Thus the points on x - axis are (8, 0) and (-2, 0).