Question
Find a point on the $y-$axis which is equidistant from the points $A(6, 5)$ and $B (- 4, 3)$.
✓
Answer
We have to find a point on the $y$-axis which is equidistant from the points $A(6, 5)$ and $B (- 4, 3)$. We know that a point on $y$-axis is of the form $(0, y)$. So, let the required point be $P (0, y)$.
Then,
$PA = PB$
$\Rightarrow \sqrt { ( 0 - 6 ) ^ { 2 } + ( y - 5 ) ^ { 2 } } = \sqrt { ( 0 + 4 ) ^ { 2 } + ( y - 3 ) ^ { 2 } }$
$\Rightarrow 36+(y-5)^2=16+(y-3)^2$
$ \Rightarrow 36+y^2-10 y+25=16+y^2-6 y+9$
$ \Rightarrow 4 y=36$
$ \Rightarrow y=9$
So, the required point is $ (0, 9).$
Then,
$PA = PB$
$\Rightarrow \sqrt { ( 0 - 6 ) ^ { 2 } + ( y - 5 ) ^ { 2 } } = \sqrt { ( 0 + 4 ) ^ { 2 } + ( y - 3 ) ^ { 2 } }$
$\Rightarrow 36+(y-5)^2=16+(y-3)^2$
$ \Rightarrow 36+y^2-10 y+25=16+y^2-6 y+9$
$ \Rightarrow 4 y=36$
$ \Rightarrow y=9$
So, the required point is $ (0, 9).$
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