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JEE Main Chemistry Syllabus 2025 - Check Topic wise Syllabus

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

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Distance of a Point From a Line is considered one the most difficult concept.

  • 39 Questions around this concept.

Solve by difficulty

If the length of the perpendicular drawn from the origin to the line whose intercepts on the axes are a and b be p, then 

Concepts Covered - 1

Distance of a Point From a Line

Distance of a point from a line

Perpendicular length from a point (x1,y1) to the line L : Ax + By + C = 0 is 

\frac{\left|\mathbf{A} \mathbf{x}_{1}++\mathbf{B} \mathbf{y}_{1}+\mathbf{C}\right|}{\sqrt{\mathbf{A}^{2}+\mathbf{B}^{2}}}

Let L : Ax + By + C = 0 be a line, whose distance from the point P (x1 , y1 ) is d. Draw a perpendicular PM 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) and R\left(0,-\frac{C}{B}\right). Thus, the area of the triangle PQR is given by

\\\mathrm{area(\Delta {PQR})=\frac{1}{2} {PM} \cdot {QR} \Rightarrow {PM}=\frac{2(\text { area } \Delta {PQR})}}\\\begin{aligned} \text { also. 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 {PQR})=\left|\frac\right|\left|{Ax}_{1}+{B} y_{1}+{C}_{1}\right|\\\mathrm{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}}}\\ \text{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 ax + by + c = 0 and ax + by + d = 0, then the distance between them is the perpendicular distance of any point on one line from the other line.

If \mathrm{\left ( x_1,y_1 \right )} is any point on the line ax + by + c = 0 

Then, \mathrm{a x_{1}+b y_{1}+c=0}

Now, perpendicular distance of the point \mathrm{\left ( x_1,y_1 \right )} from the line ax + by + d = 0 is

\frac{\left|\mathbf{a x}_{1}+\mathbf{b y}_{1}+\mathbf{d}\right|}{\sqrt{\mathbf{a}^{2}+\mathbf{b}^{2}}}=\frac{|\mathbf{d}-\mathbf{c}|}{\sqrt{\mathbf{a}^{2}+\mathbf{b}^{2}}}

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Distance of a Point From a Line

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