Distance From Point To Line & Distance between Parallel Lines

Distance of Point to a Lines-

The general equation of a line is given by Ax + By + C = 0. Consider a point P in the Cartesian plane having the co-ordinates (x1,y1). The distance from point to line, in the Cartesian system is given by calculating the length of the perpendicular between the point and line.

In the figure given below, the distance between the point P and the line LL can be calculated by figuring out the length of the perpendicular.

Draw PQ from P to the line L.

Distance of a point from a Line

The coordinate points for different points are as follows:

Point P (x1, y1), Point N (x2, y2), Point R (x3,y3)

The line L makes intercepts on both the x – axis and y – axis at the points N and M respectively. The co-ordinates of these points are \(M (0,-\frac{C}{B})\) and \(N~ (-\frac{C}{A},0)\).

Area of Δ MPN can be given as:

Area of Δ MPN = \( \frac{1}{2}~×~Base~×~Height\)

\(\Rightarrow Area~ of~ Δ~MPN\) = \(\frac{1}{2}~×~PQ~×~MN\)

\(\Rightarrow PQ\) = \(\frac{2~×~Area~ of~ Δ~MPN}{MN}\)   ………………………(i)

In terms of Co-ordinate Geometry, the area of the triangle is given as:

Area of Δ MPN = \( \frac{1}{2} \left [ x_{1} (y_{2}-y_{3}) + x_{2} (y_{3}-y_{1}) + x_{3} (y_{1}-y_{2})\right ]\)

Therefore, the area of the triangle can be given as:

Area of Δ MPN \(= \frac{1}{2} \left [ x_{1} (0 + \frac{C}{B}) + (-\frac{C}{A}) ( -\frac{C}{B} -y_{1}) + 0( y_{1}-0 )\right ]\)

\(\Rightarrow Area ~of~ Δ~MPN\)  \(= \frac{1}{2} \left [\frac{C}{B} \times x_{1} + \frac{C}{A} \times y_{1} + (\frac{c^{2}}{AB}))\right ]\)

Solving this expression we get;

\(2~×~Area~ of~ Δ~MPN\) \(= \left ( \frac{C}{AB} \right ) (Ax_{1} + B y_{1} + C)\)   …………………………(ii)

Using the distance formula, we can find out the length of the side MN of ΔMPN.

\(MN = \sqrt{\left ( 0 + \frac{C}{A} \right )^{2} + \left ( \frac{C}{B}- 0 \right )^{2}}\)

\(\Rightarrow MN = \frac{C}{AB} \sqrt{A^{2} + B^{2}}\)   …………………………………..(iii)

Equating equation (ii) and (iii) in (i), the value of perpendicular comes out to be:

\(PQ\) \(= \frac{\left | Ax_{1} + By_{1} + C \right |}{\sqrt{A^{2} + B^{2}}}\)

This length is generally represented by \(d\).

Distance Between Two Parallel Lines-

The distance between two parallel lines is equal to the perpendicular distance between the two lines. We know that, slopes of two parallel lines are same; therefore the equation of two parallel lines can be given as:

\(y\) = \(mx~ + ~c_1\) and \(y\) = \(mx ~+ ~c_2\)

The point \(A\) is the intersection point of the second line on the \(x\) – axis.

Distance Between two Parallel Lines

The perpendicular distance would be the required distance between two lines

The distance between the point \(A\) and the line \(y\) = \(mx ~+ ~c_2\) can be given by using the formula:

\(d\) = \(\frac{\left | Ax_{1} + By_{1} + C \right |}{\sqrt{A^{2} + B^{2}}}\)

\(\Rightarrow d \) \(= \frac{\left | (-m)(\frac{-c_{1}}{m}) –  c_{2} \right |}{\sqrt{1 + m^{2}}}\)

\(\Rightarrow d \) \(= \frac{\left | c_{1} – c_{2} \right |}{\sqrt{1 + m^{2}}}\)

Thus we can conclude that the distance between two parallel lines is given by:

\(d\) = \(\frac{|c_1 ~- ~c_2|}{√1 + m^2}\)

If we consider the general form of the equation of straight line, and the lines are given by:

\( L_1 ~: ~Ax~ + ~By ~+ ~C_1\) = \(0\)

\( L_2 ~: ~Ax ~+ ~By ~+ ~C_2\) = \(0 \)

Then, the distance between them is given by:

\(d\) = \(\frac{|C_1 ~- ~C_2|}{√A^2~ +~ B^2}\)<

Thus we can now easily calculate the distance between two parallel lines and the distance between a point and a line. To know more about co-ordinate geometry stay connected with us. Join us at www.byjus.com to know more.


Practise This Question

Which of the following shows  12 and 112 on the number line?