Here you will learn formula to find the distance of a point from a line with examples.
Let’s begin –
Distance of a Point from a Line
The length of the perpendicular from a point \((x_1, y_1)\) to a line ax + by + c = 0 is
|\(ax_1 + by_1 + c\over {\sqrt{a^2+b^2}}\)|.
It is the distance of a point from a line.
Distance of a Line from Origin
The length of the perpendicular from the origin to a line ax + by + c = 0 is
\( | c |\over {\sqrt{a^2+b^2}}\).
Algorithm to find distance :
Step 1 : Write the equation of the line in the form ax + by + c = 0
Step 2 : Substitute the coordinates \(x_1\) and \(y_1\) of the point in place of x and y respectively in the expression.
Step 3 : Divide the result obtained in step 2 by the square root of the sum of the squares of the coefficients of x and y.
Step 4 : Take the modulus of the expression obtained in step 3.
he result obtained after step 4 is the required distance.
Example : Find the distance between the line 12x – 5y + 9 = 0 and the point (2,1).
Solution : We have line 12x – 5y + 9 = 0 and the point (2,1)
Required distance = |\(12*2 – 5*1 + 9\over {\sqrt{12^2 + (-5)^2}}\)|
= \(|24-5+9|\over 13\) = \(28\over 13\)
Example : If p is the length of the perpendicular from the origin to the line \(x\over a\) + \(y\over b\) = 1, then prove that \(1\over p^2\) = \(1\over a^2\) + \(1\over b^2\)
Solution : The given line is bx + ay – ab = 0 ………….(i)
It is given that
p = Length of the perpendicular from the origin to line (i)
\(\implies\) p = \(|b(0) + a(0) – ab|\over {\sqrt{b^2+a^2}}\) = \(ab\over \sqrt{a^2+b^2}\)
\(\implies\) \(p^2\) = \(a^2b^2\over a^2+b^2\) \(\implies\) \(1\over p^2\) = \(a^2+b^2\over a^2b^2\) \(\implies\) \(1\over p^2\) = \(1\over a^2\) + \(1\over b^2\)
Hence Proved.