Find Perpendicular Distance between a Point and the Line

QUESTION: What is the length of the perpendicular distance from point (-2, -3) to the line 5x-2y+4=0?

SOLUTION:

The standard form of a line is

ax + by + c = 0

where x, y are the variables and a, b, c are the constants such that not both a and b are zero.

The formula for finding the perpendicular distance `d` between point $(x_1,\;y_1)$ and the line ax + by + c = 0 is

$d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$ ————(A)

Here

$(x_1,y_1)$ = (-2, -3)

which implies

$x_1$ = – 2

and

$y_1$ = – 3

Compare the given equation of line 5x – 2y + 4 = 0 with the standard equation of line ax + by + c = 0 we get

a = 5

b = -2

c = 4

Put values of $x_1$ , $y_1$ , a, b, c in equation (A)

$d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$

$d=\frac{|5(-2)+(-2)(-3)+4|}{\sqrt{5^2+(-2)^2}}$

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

$d=\frac{|-10+10|}{\sqrt{25+4}}$

$d=\frac0{\sqrt{29}}$

$d=0$

Since the perpendicular distance between the point (-2, -3) and the line 5x-2y+4=0 is zero, so the point (-2, -3) lies on the line 5x-2y+4=0.

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