Answer:
5 units
Step-by-step explanation:
You want the distance from P(-1, 5) to the line through points (0, -2) and (6, 6).
Line
The equation of the line can be written as ...
(y2 -y1)(x -x1) -(x2 -x1)(y -y1) = 0
(6 -(-2))(x -0) -(6 -0)(y -(-2)) = 0
8x -6(y +2) = 0
8x -6y -12 = 0
These coefficients are all divisible by 2, so we can reduce this to ...
4x -3y -6 = 0
Distance
The distance from point (x, y) to line ax+by+c = 0 is given by ...
[tex]d=\dfrac{|ax+by+c|}{\sqrt{a^2+b^2}}[/tex]
For the given point and line, the distance is ...
[tex]d=\dfrac{|4(-1)-3(5)-6|}{\sqrt{4^2+(-3)^2}}=\dfrac{|-25|}{\sqrt{25}}=\dfrac{25}{5}=\boxed{5}[/tex]
The distance from the point to the line is 5 units.
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Additional comment
There are several other ways you can go about this. The attached graph demonstrates a graphical solution could work nicely. The line between the given points has slope 4/3, so the perpendicular line through (-1, 5) will have slope -3/4. The point of intersection of the two lines is the midpoint between the points on the given line. The distance of interest is the hypotenuse of a 3-4-5 right triangle, so is 5 units.
