Question

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

• A. $\frac{23}{7}\sqrt{5}$ sq units
• B. $\frac{23}{18}\sqrt{5}$ sq units
• C. $\frac{23}{8}\sqrt{5}$ sq units
• D. None of these

Answer 1

Answer: (B)

The equation of line $AB$ is
$4x+7y+5=0$ ...(i)

and equation of line $PQ$ is
$2x−y=0$ ... (ii)
On solving Eqs. (i) and (ii), we get
$x=−\frac{5}{18}$ and $y=−\frac{5}{9}$
$∴$ Coordinates of $Q=\left(−\frac{5}{18},−\frac{5}{9}\right)$
Length of $QP=\sqrt{{\left(x_2−x_1\right)}^2+{\left(y_2−y_1\right)}^2}$
$=\sqrt{{\left(1+\frac{5}{18}\right)}^2+{\left(2+\frac{5}{9}\right)}^2}$
$=\sqrt{{\left(\frac{23}{18}\right)}^2+{\left(\frac{23}{9}\right)}^2}$
$=\frac{23}{9}\sqrt{\frac{1}{4}+1}=\frac{23}{9}×\frac{\sqrt{5}}{2}$
$=\frac{23}{18}\sqrt{5}sq$ units