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Question 1181250: Given triangle OAB with vertices O(0, 0), A(5, 2), and B(a, b), let AP and BQ be the perpendiculars dropped from A and B to their opposite sides of triangle OAB ( a not equal to 0, b not equal to 0). Find the constants a and b when the orthocenter of triangle OAB is H(3, 1).
P.S. Please show full solution, ty.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The orthocenter of a triangle is the point of intersection of the three altitudes of the triangle. The problem will be solved making repeated use of the equations of perpendicular lines.
We have O(0,0) and A(5,2), so the equation of the line containing side OA is y=(2/5)x.
Altitude BQ passes through P(3,1) and is perpendicular to OA, which has slope 2/5; so the line containing altitude BQ has slope -5/2. Algebra shows BQ is contained in the line y=(-5/2)x+17/2.
Altitude AP passes through A(5,2) and H(3,1); easy algebra shows AP is contained in the line y=(1/2)x-1/2, which has slope 1/2.
Side OB passes through (0,0) and is perpendicular to AP, so its slope is -2; the equation of side OB is then y=-2x.
Point B is the intersection of altitude BQ and side OB:
ANSWER: B(17,-34)
To confirm our method and our calculations, we can show that the altitude from O to side AB gives an equation for the line containing side AB that gives us the same coordinates for B.
The slope of the altitude from O is 1/3, so the slope of AB is -3.
AB with a slope of -3 and passing through A(5,2) gives the equation y=-3x+17 for the line containing side AB.
Substitution verifies that (17,-34) is on that line.
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