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

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

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

1. $$\huge \underline \bold \blue{AnSwEr ♡}$$

distances are equal from both the points so

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

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

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

After solving this we get:-

y = 19.5

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