Given points A(3, 4) and B(5, -2). Find the p | vedclass

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(C) $1$. Since $PA = PB$,point $P(x, y)$ lies on the perpendicular bisector of segment $AB$.$2$. The midpoint of $AB$ is $M = (\frac{3+5}{2}, \frac{4-

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Both ($7$,$2$) and ($1$,$0$)

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(C) $1$. Since $PA = PB$,point $P(x, y)$ lies on the perpendicular bisector of segment $AB$.$2$. The midpoint of $AB$ is $M = (\frac{3+5}{2}, \frac{4-2}{2}) = (4, 1)$.$3$. The slope of $AB$ is $m_{AB} = \frac{-2-4}{5-3} = \frac{-6}{2} = -3$.$4$. The slope of the perpendicular bisector is $m_{\perp} = -\frac{1}{-3} = \frac{1}{3}$.$5$. The equation of the perpendicular bisector is $y - 1 = \frac{1}{3}(x - 4) \implies 3y - 3 = x - 4 \implies x = 3y + 1$.$6$. The area of $\Delta PAB = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 10$.$7$. Substituting $P(3y+1, y)$,$A(3, 4)$,and $B(5, -2)$: $10 = \frac{1}{2} |(3y+1)(4 - (-2)) + 3(-2 - y) + 5(y - 4)|$.$8$. $20 = |(3y+1)(6) - 6 - 3y + 5y - 20| = |18y + 6 - 26 + 2y| = |20y - 20|$.$9$. $|20y - 20| = 20 \implies |y - 1| = 1$.$10$. Case $1: y - 1 = 1 \implies y = 2$. Then $x = 3(2) + 1 = 7$. Point $P = (7, 2)$.$11$. Case $2: y - 1 = -1 \implies y = 0$. Then $x = 3(0) + 1 = 1$. Point $P = (1, 0)$.

Given points $A(3, 4)$ and $B(5, -2)$. Find the point $P(x, y)$ on the plane such that $PA = PB$ and the area of $\Delta PAB = 10$.

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