Question

Find the point on y axis which is equidistant from the points (3,2) and (4,6)

Answer 1

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distances are equal from both the points so

$\sqrt{(x1-x)^{2}+(y1-y)^2 } = \sqrt{(x2-x)^{2}+(y2-y)^2 }(x1−x)2+(y1−y)2=(x2−x)2+(y2−y)2</p><p>\sqrt{(3-x)^{2}+(2-y)^2 } = \sqrt{(4-x)^{2}+(6-y)^2 }(3−x)2+(2−y)2=(4−x)2+(6−y)2$

as the point is in y-axis then x=0, so

$\sqrt{(3)^{2}+(2-y)^2 } = \sqrt{(4)^{2}+(6-y)^2 }(3)2+(2−y)2=(4)2+(6−y)2</p><p>\ (3)^{2}+(2-y)^2 = \ (4)^{2}+(6-y)^2 (3)2+(2−y)2= (4)2+(6−y)2$

After solving this we get:-

y = 19.5

• so the final coordinate is (0, 19.5).

Answer 2

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