Question

find square point on the x-axis which is equidistant from (2,-5) and (-2,9)

Answer 1

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$<i>$
Let the point of x-axis be P(x, 0) 

Given A(2, -5) and B(-2, 9) are equidistant from P 

That is PA = PB 

Hence PA2 = PB2  → (1)Distance between two points is √[(x2 - x1)2 + (y2 - y1)2] 

PA = √[(2 - x)2 + (-5 - 0)2] 

PA2 = 4 - 4x +x2 + 25 = x2 - 4x + 29

Similarly, PB2 = x2 + 4x + 85

 Equation (1) becomes x2 - 4x + 29 = x2 + 4x + 85 - 8x = 56 x = -7 

Hence the point on x-axis is (-7, 0)

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