Question

A line contains points (−2,−2) and (1,4). Find the distance between the line and the point (6,−1).

Answer 1

In analytical geometry, derived equations are already available for finding the distance between a point and a line. The equation is


`d = \fracAx + By+ C{ \sqrt{ A^(2)+ B^(2) } }`

This is basing on the standard equation of a line in the form of:
Ax + By + C = 0, where A, B and C are coefficients.
These constants are coming from the equation of the line, while the point (x,y) is substituted to the x and y terms of the equation.

However, we must first know the equation of the line. Let us use the slope-intercept form: y = mx + b, where m is the slope which is equal to (y2-y1)/(x2-x1) while b is the y-intercept. For points (−2,−2) and (1,4), the slope is

m = (4--2)/(1--2) = 2

Then, we substitute any of those points to determine b. The answer would be the same either way. Let's use (-2,-2).
-2 = 2(-2) + b
b = 2

Thus, the equation of the line is y = 2x + 2. Rearranging into the standard form, it becomes 2x - y + 2 = 0. Therefore, A=2, B=-1 and C=2. Substituting the values to the equation by replacing x and y with coordinates of point (6,−1),


`d = \frac2(6) + (-1)(-1)+ 2{ \sqrt{ 2^(2)+ (-1)^(2) } }`

d = 3√5 or 6.71 units