Find a point on the y-axis …
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Posted by Soumyadeep Dhali 4 years, 10 months ago
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Sia ? 4 years, 10 months ago
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
{tex}\Rightarrow \sqrt { ( 0 - 6 ) ^ { 2 } + ( y - 5 ) ^ { 2 } } = \sqrt { ( 0 + 4 ) ^ { 2 } + ( y - 3 ) ^ { 2 } }{/tex}
{tex}\Rightarrow{/tex} 36 + (y - 5)2 = 16 + (y - 3)2
{tex}\Rightarrow{/tex} 36 + y2 - 10y + 25 = 16 + y2 - 6y + 9
{tex}\Rightarrow{/tex} 4y = 36
{tex}\Rightarrow{/tex} y = 9
So, the required point is (0, 9).
0Thank You