Distance of a Point From a Line - Practice Questions & MCQ

Distance of a point from a line

Perpendicular length from a point $\left(x_1, y_1\right)$ to the line $L$ : $A x+B y+C=0$ is

$
\frac{\left|\mathbf{A x}_1++\mathbf{B} \mathbf{y}_1+\mathbf{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}
$
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 $L$

The line meets the x -and y -axes at the points Q and R , respectively. Then, the coordinates of the points are $Q\left(-\frac{C}{A}, 0\right) \quad R\left(0,-\frac{C}{B}\right)$.Thus, the area of the triangle PQR is given by

$
\begin{aligned}
& \text { area }(\triangle \mathrm{PQR})=\frac{1}{2} \mathrm{PM} \cdot \mathrm{QR} \Rightarrow \mathrm{PM}=\frac{2(\operatorname{area} \Delta \mathrm{PQR})}{\mathrm{QR}} \\
& \text { also. } \begin{aligned}
\text { area }(\Delta \mathrm{PQR}) & =\frac{1}{2}\left|\mathrm{x}_1\left(0+\frac{\mathrm{C}}{\mathrm{~B}}\right)+\left(-\frac{\mathrm{C}}{\mathrm{~A}}\right)\left(-\frac{\mathrm{C}}{\mathrm{~B}}-\mathrm{y}_1\right)+0\left(\mathrm{y}_1-0\right)\right| \\
& =\frac{1}{2}\left|\mathrm{x}_1 \frac{\mathrm{C}}{\mathrm{~B}}+\mathrm{y}_1 \frac{\mathrm{C}}{\mathrm{~A}}+\frac{\mathrm{C}^2}{\mathrm{AB}}\right|
\end{aligned} \\
& \text { or } 2 \times \operatorname{area}(\Delta P Q R)=\left|\frac{C}{A B}\right|\left|A x_1+B y_1+C_1\right|
\end{aligned} \begin{aligned}
& \mathrm{QR}=\sqrt{\left(0+\frac{\mathrm{C}}{\mathrm{~A}}\right)^2+\left(\frac{\mathrm{C}}{\mathrm{~B}}-0\right)^2}=\left|\frac{\mathrm{C}}{\mathrm{AB}}\right| \sqrt{\mathrm{A}^2+\mathrm{B}^2}
\end{aligned}
$
Substituting the values

$
\mathrm{PM}=\frac{\left|\mathrm{Ax}_1++\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}
$

Distance between two parallel lines

The equation of two parallel line is $a x+b y+c=0$ and $a x+b y+d=0$, then the distance between them is the perpendicular distance of any point on one line from the other line.
If $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is any point on the line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$
Then, $\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}=0$
Now, perpendicular distance of the point $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ from the line $a x+b y+d=0$ is

$
\frac{\left|\mathrm{ax}_1+\mathrm{by}_1+\mathrm{d}\right|}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}=\frac{|\mathrm{d}-\mathrm{c}|}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}
$