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

=  \(\frac{4}{\sqrt{1+9}}\),

= \(\frac{4}{\sqrt{10}}\).