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Question 389312: Hello, I can't seem to figure this out and was hoping someone could help.
Find the coordinates of point D on line BC such that AD is the shortest distance from A to BC for A(9,9, B(-1,-1), and C(-2,2).
Thanks.
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! The slope from the point B(-1,-1) to C (-2,2) is (2--1)/(-2--1) = 3/(-1) = -3. The equation of the line is y--1 = -3(x--1) ----->y+1 = -3(x+1) ----->y = -3x-3 - 1 -----> y = -3x-4.
Now the find the equation of the line perpendicular to the line y = -3x-4 and passing through A(9,9). The slope of that line must be 1/3 (the negative reciprocal of -3). Then the equation of the line is y - 9 = (x-9)/3.
----> 3y - 27 = x - 9 ----> 3y = x + 18.
We now have to solve the system
y = -3x-4
3y = x + 18
Substitute the top equation into the bottom equation:
3(-3x-4) = x + 18;
-9x-12 = x + 18;
-30 = 10x;
x = -3.------>y = -3*-3 - 4 = 9-4 = 5.
Therefore segment AD is minimum for the point D(-3,5).
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