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.

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.

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:

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