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Q.
The perpendicular distance $(d)$ of the line $A x+B y+C=0$ from a point $\left(x_1, y_1\right)$ is given by $d=\frac{\left|A x_1+B y_1+C\right|}{m}$, where $m$ stands for
Straight Lines
Solution:
Let $L: A x+B y+C=0$ be a line, whose distance from the point $P\left(x_1, y_1\right)$ is $d$. Draw a perpendicular $P M$ 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 $\triangle P Q R$ is given by $\operatorname{area}(\triangle P Q R)=\frac{1}{2} P M \cdot Q R$, which gives
$P M=\frac{2 \text { area }(\triangle P Q R)}{Q R}$.... (i)
Also, area $(\triangle P Q R)$
$=\frac{1}{2}\left|x_1\left(0+\frac{C}{B}\right)+\left(-\frac{C}{A}\right)\left(-\frac{C}{B}-y_1\right)+0\left(y_1-0\right)\right|$
$ =\frac{1}{2}\left|x_1 \frac{C}{B}+y_1 \frac{C}{A}+\frac{C^2}{A B}\right| $
or $ 2 $ area $(\triangle P Q R)=\left|\frac{C}{A B}\right| \cdot\left|A x_1+B y_1+C\right|$
and $ Q R=\sqrt{\left(0+\frac{C}{A}\right)^2+\left(\frac{C}{B}-0\right)^2} $
$ =\left|\frac{C}{A B}\right| \sqrt{A^2+B^2} $
Substituting the values of area $(\triangle P Q R)$ and $Q R$ in Eq. (i), we get
$ P M=\frac{\left|A x_1+B y_1+C\right|}{\sqrt{A^2+B^2}} $
or $ d=\frac{\left|A x_1+B y_1+C\right|}{\sqrt{A^2+B^2}}$
Thus, the perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $\left(x_1, y_1\right)$ is given by
$d=\frac{\left|A x_1+B y_1+C\right|}{\sqrt{A^2+B^2}}$
