Suppose P and Q are the midpoints of the side | vedclass

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(A) Given,$P$ and $Q$ are the midpoints of sides $AB$ and $BC$ of a triangle with vertices $A(1, 3)$,$B(3, 7)$,and $C(7, 15)$.Coordinates of $P = \lef

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$6x + 12y = 297$

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(A) Given,$P$ and $Q$ are the midpoints of sides $AB$ and $BC$ of a triangle with vertices $A(1, 3)$,$B(3, 7)$,and $C(7, 15)$.Coordinates of $P = \left(\frac{1+3}{2}, \frac{3+7}{2}\right) = (2, 5)$.Coordinates of $Q = \left(\frac{3+7}{2}, \frac{7+15}{2}\right) = (5, 11)$.Let the coordinates of $R$ be $(x, y)$.The given condition is $AC^2 + QR^2 = PR^2$,which implies $PR^2 - QR^2 = AC^2$.$AC^2 = (7-1)^2 + (15-3)^2 = 6^2 + 12^2 = 36 + 144 = 180$.$PR^2 - QR^2 = [(x-2)^2 + (y-5)^2] - [(x-5)^2 + (y-11)^2] = 180$.$(x^2 - 4x + 4 + y^2 - 10y + 25) - (x^2 - 10x + 25 + y^2 - 22y + 121) = 180$.$(6x + 12y - 117) = 180$.$6x + 12y = 297$.

Suppose $P$ and $Q$ are the midpoints of the sides $AB$ and $BC$ of a triangle where $A(1, 3)$,$B(3, 7)$,and $C(7, 15)$ are vertices. Then the locus of $R$ satisfying $AC^2 + QR^2 = PR^2$ is

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