Find the point on the Y-axis which is equidis | vedclass

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Question

(A) Let the point on the $Y$-axis be $P(0, y)$.Since $P$ is equidistant from $A(3, 1)$ and $B(-2, 5)$, we have $PA = PB$, which implies $PA^2 = PB^2$.

Vedclass · Accepted Answer

$(0, \frac{19}{8})$

Vedclass · Answer

(A) Let the point on the $Y$-axis be $P(0, y)$.Since $P$ is equidistant from $A(3, 1)$ and $B(-2, 5)$, we have $PA = PB$, which implies $PA^2 = PB^2$.Using the distance formula: $(3 - 0)^2 + (1 - y)^2 = (-2 - 0)^2 + (5 - y)^2$.$9 + (1 - 2y + y^2) = 4 + (25 - 10y + y^2)$.$10 - 2y + y^2 = 29 - 10y + y^2$.Subtracting $y^2$ from both sides: $10 - 2y = 29 - 10y$.Rearranging terms: $10y - 2y = 29 - 10$.$8y = 19$.$y = \frac{19}{8}$.The point is $(0, \frac{19}{8})$.

Find the point on the $Y$-axis which is equidistant from $(3, 1)$ and $(-2, 5)$.

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