- #1
Imperil
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In a backyard, there are two trees located at grid points A(-2, 3) and B(4, -6). The family cat is walking in the backyard. The line segments between the cat and the two trees are always perpendicular. Find the equation of the locus of the cat.
My Answer:
slope PA = (y - 3) / (x + 2)
slope PB = (y + 6) / (x - 4)
slope PA = -1 / slope PB (since the lines between PA and PB are perpendicular)
(y - 3) / (x + 2) = -1 / [(y + 6) / (x - 4)]
(y - 3) / (x + 2) = -1 * (x - 4) / (y + 6)
(y - 3) / (x + 2) = (-x+4) / (y + 6)
(y - 3) (y + 6) = (x + 2)(-x + 4)
y^2 + 3y - 18 = -x^2 + 2x + 8
x^2 + y^2 - 2x + 3y - 26 = 0
x^2 - 2x + 1 + y^2 + 3y + 2.25 = 26 + 1 + 2.25
(x - 1)^2 + (y + 1.5)^2 = 29.25
Therefore the cat is walking in a circle with center (1, -1.5) and radius sqrt(29.25).
I believe that my answer is incorrect but is there something I am missing? I have tried doing this question multiple times and I still can't find the correct answer.
EDIT: corrected a typo
My Answer:
slope PA = (y - 3) / (x + 2)
slope PB = (y + 6) / (x - 4)
slope PA = -1 / slope PB (since the lines between PA and PB are perpendicular)
(y - 3) / (x + 2) = -1 / [(y + 6) / (x - 4)]
(y - 3) / (x + 2) = -1 * (x - 4) / (y + 6)
(y - 3) / (x + 2) = (-x+4) / (y + 6)
(y - 3) (y + 6) = (x + 2)(-x + 4)
y^2 + 3y - 18 = -x^2 + 2x + 8
x^2 + y^2 - 2x + 3y - 26 = 0
x^2 - 2x + 1 + y^2 + 3y + 2.25 = 26 + 1 + 2.25
(x - 1)^2 + (y + 1.5)^2 = 29.25
Therefore the cat is walking in a circle with center (1, -1.5) and radius sqrt(29.25).
I believe that my answer is incorrect but is there something I am missing? I have tried doing this question multiple times and I still can't find the correct answer.
EDIT: corrected a typo
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