The point on the Y- axis which is equidistant | vedclass

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Question

(B) Suppose the required point on the $Y-$ axis is $R(0, y)$.Since the point is equidistant from $P(3, 2)$ and $Q(-1, -5)$,we have $PR = QR$.Squaring

Vedclass · Accepted Answer

$\left(0,-\frac{13}{14}\right)$

Vedclass · Answer

(B) Suppose the required point on the $Y-$ axis is $R(0, y)$.Since the point is equidistant from $P(3, 2)$ and $Q(-1, -5)$,we have $PR = QR$.Squaring both sides,we get $PR^2 = QR^2$.Using the distance formula,$(3 - 0)^2 + (2 - y)^2 = (-1 - 0)^2 + (-5 - y)^2$.Expanding the squares,$9 + (4 - 4y + y^2) = 1 + (25 + 10y + y^2)$.Simplifying the equation,$13 - 4y + y^2 = 26 + 10y + y^2$.Subtracting $y^2$ from both sides,$13 - 4y = 26 + 10y$.Rearranging the terms,$-13 = 14y$.Therefore,$y = -\frac{13}{14}$.The required point on the $Y-$ axis is $\left(0, -\frac{13}{14}\right)$.

The point on the $Y-$ axis which is equidistant from $P(3,2)$ and $Q(-1,-5)$ is $\ldots \ldots \ldots \ldots$

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