Question

Which point on y-axis is equidistant from the points (12,3) and (-5,10)

Answer 1

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Answer 2

The point on the y-axis which is equidistant from points (12. 3) and (-5, 10) is (0, -2).

Step-by-step Explanation:

Given: Two points, (12. 3) and (-5, 10)

To be found: To find the point on the y-axis which is equidistant from the given two points.

Formula Used: The distance between two points, $(x_{1} ,y_{1} )$ and $(x_{2} ,y_{2} )$ is calculated by the formula,

$Distance=\sqrt{(x_{2} -x_{1} )^{2} +(y_{2} -y_{1})^{2} }$

Solution:

We have the points, A(12, 3) and B(-5, 10).

Let C(0, y) be a point on the y-axis which is equidistant from the given two points.

$\implies CA=CB$  -----------(1)          (Since C is equidistant from A and B)

Calculating the distance of CA and CB using the distance formula, (1) becomes

$\sqrt{(12 -0 )^{2} +(3-y)^{2} } = \sqrt{(-5-0 )^{2} +(10-y)^{2} } \\\implies \sqrt{(12 )^{2} +(9+y^{2}-6y) }= \sqrt{(-5 )^{2} +(100+y^{2}-20y) }$  

$[\because (a-b)^{2}=a^{2} +b^{2} -2ab]$

Squaring on both sides, we get

$144 +9+y^{2}-6y=25+100+y^{2} -20y\\\implies 153+y^{2}-6y=125+y^{2} -20y$

Taking the terms with y to one side, we get

$y^{2} -y^{2} -6y+20y=125-153\\\implies 14y=-28$

Simplifying, we get

$y=-2$

Therefore, (0, -2) is the point on the y-axis that is equidistant from the given two points.

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