1.

Which point on the y-axis is equidistant from (2, 3) and (-4, 1)?

Answer»

Let A (2, 3) and B (-4, 1) be the given points

Let the point on y – axis equidistant from the above points be C (0, y)

Now, we have

AC = √[(0 – 2)2 + (y – 3)2

= √[y2 – 6y + 9 + 4] 

= √[y2 – 6y + 13]

And,

BC = √[(0 – (-4))2 + (y – 1)2

= √[y2 – 2y + 1 + 16] 

= √[y2 – 2y + 17]

As AC = BC (given condition)

So, AC2 = BC2

y2 – 6y + 13 = y2 – 2y + 17

-4y = 4

y = -1

Therefore, the point on the y-axis is (0, -1).



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