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Question 1197253: find the distance of the point (12,7)from the line through (5,3), (2,-3)
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Let's find the slope of the line through (5,3) and (2,-3)
(x1,y1) = (5,3)
(x2,y2) = (2,-3)
m = slope
Now apply point-slope form
y-y1 = m(x - x1)
y-3 = 2(x - 5)
y-3 = 2x - 10
y = 2x - 10+3
y = 2x - 7
This is in y = mx+b form
Let's get everything to one side
y = 2x - 7
0 = 2x - 7 - y
0 = 2x - y - 7
2x-y-7 = 0
This is now in the form Ax+By+C=0 with A = 2, B = -1, C = -7
Now we'll use the formula discussed here
https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line
I'll use slightly different notation
The A,B,C will be the same
Instead of (x0,y0) I'll refer to the point not on the line as (p,q)
The distance from (p,q) to the line Ax+By+C = 0 is  where point (p,q) is not on the line mentioned.
We'll plug in A = 2, B = -1, C = -7
And also p = 12 and q = 7 from the point (12,7)
 Multiply top and bottom by sqrt(5) to rationalize the denominator
 is the exact distance from (12,7) to the line 2x-y-7 = 0 which goes through (5,3) and (2,-3)
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Verification:
I used GeoGebra to make this image.
The red line 2x-y-7 = 0 was what we found earlier through (5,3) and (2,-3)
The black line x+2y-26 = 0 is perpendicular to the red line, and it goes through point A(12,7)
Use whichever method you prefer to solve that system of 2 equations to find the red and black line intersect at D(8,9)
Then use the distance formula to find the distance from A(12,7) to D(8,9) is exactly  units.
Side note:
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