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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