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,
$PA = PB$
$\Rightarrow \sqrt { ( 0 - 6 ) ^ { 2 } + ( y - 5 ) ^ { 2 } } = \sqrt { ( 0 + 4 ) ^ { 2 } + ( y - 3 ) ^ { 2 } }$
$\Rightarrow$ $36 + (y - 5)^2 = 16 + (y - 3)^2$
$\Rightarrow$ $36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9$
$\Rightarrow$ $4y = 36$
$\Rightarrow$ $y = 9$
So, the required point is (0, 9).
Then,
$PA = PB$
$\Rightarrow \sqrt { ( 0 - 6 ) ^ { 2 } + ( y - 5 ) ^ { 2 } } = \sqrt { ( 0 + 4 ) ^ { 2 } + ( y - 3 ) ^ { 2 } }$
$\Rightarrow$ $36 + (y - 5)^2 = 16 + (y - 3)^2$
$\Rightarrow$ $36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9$
$\Rightarrow$ $4y = 36$
$\Rightarrow$ $y = 9$
So, the required point is (0, 9).
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