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Question 1043695: P and Q are two points on the line x-y+1=0 and are at the distance 5 units from origin.Find the area of the triangle OPQ.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! P and Q are two points on the line x-y+1=0 and are at the distance 5 units from origin.Find the area of the triangle OPQ.
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x-y+1 = 0
y = x+1
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Slope = 1
atan(1) = 45 degrees
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P and Q are two points on the line x-y+1=0 and are at the distance 5 units from origin.
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bb later to finsih. The plumber is here.
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find the 2 points P and Q
It's the intersections of the line and the circle.
y = x+1
x^2 + y^2 = 5^2
x^2 + (x+1)^2 = 25
2x^2 + 2x - 24 = 0
x^2 + x - 12 = 0
(x+4)*(x-3) = 0
x = -4, y = -3 --> (-4,-3)
x = 3, y = 4 --> (3,4)
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Pq = sqrt(diffy^2 + diffx^2) = 7sqrt(2)
It's a triangle with sides of 5, 5 and 7sqrt(2)
Use Heron's Law to find the area.
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Or,
Use point R at (3,-3) to from a right triangle PQR.
The area of PQR is 7*7/2 = 49/2 sq units.
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Subtract the area of triangle OQ(3,0) and the area of OP(0,-3), and subtract the area of the square. The square is 3 by 3 --> area = 9 sq units.
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