Calculate the distance between a point and a line

Question

Calculate the distance between point P and the line r, in the examples below

P =(1,4), r=3x-4y-7 = 0

P =(2,-3), r=5x + 12y -39=0

Answer 1

P =(1,4), r=3x-4y-7 = 0

First, put the line in the "slope-intercept" format:

3x - 4y - 7 = 0

-4y = -3x + 7

y = 3x/4 - 7/4 (original line)

The (shortest) distance from this line to a given point is along a perpendicular line. This will have a slope of -4/3 (the negative reciprocal of the original line)

Use this new slope and the given point to find the equation of the new perpendicular line:

y = mx + b

4 = -4/3 * 1 + b

Solve for b:

b = 4 + 4/3 = 16/3

So, the new line is:

y = -4x/3 + 16/3 (new line, perpendicular to original, and passing through (1,4)

Next, find the point of intersection

Subtract the new equation from the original to get:

0 = 3x/4 - 7/4 - (-4x/3 + 16/3 )

3x/4 + 4x/3 = 7/4 + 16/3

9x + 16x = 21 + 64

25x = 85

x = 17/5

y = -4/3 * 17/5 + 16/3

= -68/15 + 80/15 = 12/15

y = 4/5 (intersection is (17/5, 4/5)

Finally, find the distance by finding the differences in x and y and applying the Pythagorean Theorem