# Perpendicular Distance from a Point to the Line – Problem

### Perpendicular Distance from a Point to the Line – Problem

Formula: The perpendicular distance from a point p (x, y) to the line ax + by + c = 0 is $$\frac{|a{{x}_{1}}+b{{y}_{1}}+c|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$$.

1) 5x – 2y + 4 = 0. (-2, -3)

Solution: Given that 5x – 2y + 4 = 0 … (1)

Length of the perpendicular (d) = $$\frac{|a{{x}_{1}}+b{{y}_{1}}+c|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$$,

(x₁, y₁) = (-2, -3)

Now d =$$\frac{|5x-2y+4|}{\sqrt{{{5}^{2}}+{{(-2)}^{2}}}}$$,

$$\frac{|5(-2)-2(-3)+4|}{\sqrt{25+4}}$$,

$$\frac{|-10+6+4|}{\sqrt{25+4}}$$,

$$\frac{|0|}{\sqrt{25+4}}$$.

The perpendicular distance from a point (-2, -3) to the line (1) = 0

2) x – 3 y – 4 = 0. (0, 0)

Solution: Length of the perpendicular (d) = $$\frac{|c|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$$.

(x₁, y₁) = (0, 0)

Now d =  $$\frac{|-4|}{\sqrt{{{1}^{2}}+{{(-3)}^{2}}}}$$,

=  $$\frac{4}{\sqrt{1+9}}$$,

= $$\frac{4}{\sqrt{10}}$$.