Question

Find the distance of the line $4x + 7y + 5 = 0$ from the point (1, 2) along the line $2x - y = 0$

Answer 1

The given lines are
$2x \- y = 0 … (1)$
$4x + 7y + 5 = 0 … (2)$
A (1, 2) is a point on line (1).
Let B be the point of intersection of lines (1) and (2).

On solving equations (1) and (2), we obtain
$x =\frac{ -5}{18}$ and $y = – \frac{5}{9}$

∴Coordinates of point B are $\left(\frac{-5}{18}, \frac{-5}{9}\right).$

By using distance formula, the distance between points A and B can be obtained as
$AB=\sqrt{\left(1+\frac{5}{18}\right)^2+\left(2+\frac{5}{9}\right)^2}$units.

$=\sqrt{\left(\frac{23}{18}\right)^2+\left(\frac{23}{9}\right)^2}$ uniits

$= \sqrt{\left(\frac{23}{2\times9}\right)^2+\left(\frac{23}{9}\right)^2}$ units

$= \sqrt{\left(\frac{23}{9}\right)^2\left(\frac{1}{2}\right)^2+\left(\frac{23}{9}\right)^2}$ units

$= \sqrt{\left(\frac{23}{9}\right)^2+\left(\frac{1}{4}+1\right)}$ units

$= \frac{23}{9}\sqrt{\frac{5}{4}}$ units

$= \frac{23}{9} \times\frac{\sqrt5}{2}$units

$=\frac{23\sqrt{5}}{18}$ units

Thus, the required distance is $\frac{23\sqrt{5}}{18}$ units.