Question

3. Calculate the coordinate of the point on x‐axis which is equidistant from (2,‐5)and (‐2,9).

Answer 1

hey.. here is ur ans..    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)   #hope it helps   #Mark as the brainliest pls