Question

The perpendicular distance $(d)$ of the line $Ax+By+C=0$ from a point $\left(x_1,y_1\right)$ is given by $d=\frac{|A{x}_1+B{y}_1+C|}{m}$, where $m$ stands for

• A. $A^2+B^2$
• B. $\sqrt{A^2+B^2}$
• C. $A+B$
• D. $\sqrt{A+B}$

Answer 1

Answer: (B)

Let $L:Ax+By+C=0$ be a line, whose distance from the point $P\left(x_1,y_1\right)$ is $d$. Draw a perpendicular $PM$ from the point $P$ to the line $L$.

If the line meets the $X$ and $Y$-axes at the points $Q$ and $R$ respectively, then coordinates of the point are $Q\left(-\frac{C}{A},0\right)$ and $R\left(0,-\frac{C}{B}\right)$. Thus, the area of the $△PQR$ is given by $\mathrm{area}(△PQR)=\frac{1}{2}PM\cdot QR$, which gives
$PM=\frac{2\text{ area }(△PQR)}{QR}$.... (i)
Also, area $(△PQR)$
$=\frac{1}{2}\left|x_1\left(0+\frac{C}{B}\right)+\left(-\frac{C}{A}\right)\left(-\frac{C}{B}-y_1\right)+0\left(y_1-0\right)\right|$
$=\frac{1}{2}\left|x_1\frac{C}{B}+y_1\frac{C}{A}+\frac{C^2}{AB}\right|$
or $2$ area $(△PQR)=\left|\frac{C}{AB}\right|\cdot \left|A{x}_1+B{y}_1+C\right|$
and $QR=\sqrt{{\left(0+\frac{C}{A}\right)}^2+{\left(\frac{C}{B}-0\right)}^2}$
$=\left|\frac{C}{AB}\right|\sqrt{A^2+B^2}$
Substituting the values of area $(△PQR)$ and $QR$ in Eq. (i), we get
$PM=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$
or $d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$
Thus, the perpendicular distance $(d)$ of a line $Ax+By+C=0$ from a point $\left(x_1,y_1\right)$ is given by
$d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$