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

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Question

(C) Let the point on the $Y$-axis be $P(0, y)$.Given points are $A(-5, -2)$ and $B(3, 2)$.Since $P$ is equidistant from $A$ and $B$,we have $PA = PB$.

Vedclass · Accepted Answer

$(0, -2)$

Vedclass · Answer

(C) Let the point on the $Y$-axis be $P(0, y)$.Given points are $A(-5, -2)$ and $B(3, 2)$.Since $P$ is equidistant from $A$ and $B$,we have $PA = PB$.Using the distance formula,$PA^2 = PB^2$.$(0 - (-5))^2 + (y - (-2))^2 = (0 - 3)^2 + (y - 2)^2$$5^2 + (y + 2)^2 = (-3)^2 + (y - 2)^2$$25 + y^2 + 4y + 4 = 9 + y^2 - 4y + 4$$29 + 4y = 13 - 4y$$8y = 13 - 29$$8y = -16$$y = -2$Therefore,the required point is $(0, -2)$.

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

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