Class 10 Math All topics Coordinate Geometry

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

Solution:

Let point $P(0,y)$ is on the y-axis, then

$PA=PB$

or $PA_{2}=PB_{2}$

$(6−0)_{2}+(5−y)_{2}=(−4−0)_{2}+(3−y)_{2}$

$36+25−10y+y_{2}=16+9−6y+y_{2}$

$61−10y+25−6y$

$61−25=−6y+10y$

$36=4y$

or $y=9$

The required point is $(0,9)$.

Similar topics

introduction to trigonometry

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quadratic equations

surface areas and volumes

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