1.

Which point on x – axis is equidistant from (5, 9) and (-4, 6)?

Answer»

Let A (5, 9) and B (-4, 6) be the given points

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

Now, we have

AC = √[(x – 5)2 + (0 – 9)2

= √[x2 – 10x + 25 + 81] 

= √[x2 – 10x + 106]

And,

BC = √[(x – (-4))2 + (0 – 6)2

= √[x2 + 8x + 16 + 36] 

= √[x2 + 8x + 52]

As AC = BC (given condition)

So, AC2 = BC2

x2 – 10x + 106 = x2 + 8x + 52

18x = 54

x = 3

Therefore, the point on the x-axis is (3, 0)



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