Find a point on the y-axis which is equidista | vedclass

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Question

(A) We know that any point on the $y$-axis is of the form $(0, y)$. Let the point $P(0, y)$ be equidistant from points $A(6, 5)$ and $B(-4, 3)$.Accord

Vedclass · Accepted Answer

$(0, 9)$

Vedclass · Answer

(A) We know that any point on the $y$-axis is of the form $(0, y)$. Let the point $P(0, y)$ be equidistant from points $A(6, 5)$ and $B(-4, 3)$.According to the distance formula,$PA = PB$,so $PA^2 = PB^2$.$(6 - 0)^2 + (5 - y)^2 = (-4 - 0)^2 + (3 - y)^2$$36 + (25 - 10y + y^2) = 16 + (9 - 6y + y^2)$$61 - 10y + y^2 = 25 - 6y + y^2$Subtracting $y^2$ from both sides:$61 - 10y = 25 - 6y$$61 - 25 = 10y - 6y$$36 = 4y$$y = 9$Thus,the required point is $(0, 9)$.

Find a point on the $y$-axis which is equidistant from the points $A(6, 5)$ and $B(-4, 3)$.

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