Question

find a point on y axis which is equidistant from A(-2,3) and (5,4)​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The point on y axis which is equidistant from coordinates (-2,3) and (5,4) is 9.

Step-by-step explanation:

Let the coordinates (-2, 3) and (5, 4) be named as points A and B.

Let the C $(0, y)$ be equidistant from the given coordinates.

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

$\sqrt{[0 - (-2)]^{2} +(y -3)^{2} } = \sqrt{(0 - 5)^{2} +(y -4)^{2} }$

$\sqrt{(0+2)^{2} +(y -3)^{2} } = \sqrt{(- 5)^{2} +(y -4)^{2} }$ $($∵ $a-(-b)=a+b)$

$\sqrt{(2)^{2} +(y -3)^{2} } = \sqrt{(- 5)^{2} +(y -4)^{2} }$

$\sqrt{4+(y^{2} +9-6y) } = \sqrt{25 +(y^{2} +16-8y) }$ $($∵ $(a-b)^{2} = a^{2} +b^{2}-2ab$$)$

$\sqrt{4 +y^{2} +9-6y } = \sqrt{25+y^{2} +16-8y }$

$\sqrt{y^{2}+13-6y } = \sqrt{y^{2}+31 -8y }$

Squaring on both sides,

$y^{2}+13-6y = y^{2}+31 -8y$

$y^{2}-y^{2}+8y-6y = 31 -13$

$2y=18$

$y=\frac{18}{2}$

$y=9$

∴ The point on y axis which is equidistant from coordinates (-2,3) and (5,4) is 9.