Right answer is (c) (11, 0)
The explanation: Let the point on x-axis be (x, 0)
Distance between (9, 8) and (x, 0) = \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}
= \sqrt {(x-9)^2 + (0-8)^2}
= \sqrt {x^2-18x + 81 + (-8)^2}
= \sqrt {x^2-18x + 81 + 64}
= \sqrt {x^2-18x + 145}
Distance between (3, 2) and (x, 0) = \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}
= \sqrt {(x-3)^2 + (0-2)^2}
= \sqrt {x^2-6x + 9 + (-2)^2}
= \sqrt {x^2-6x + 9 + 4}
= \sqrt {x^2-6x + 13}
Since, the point ( x, 0) is equidistant to (3, 2) and (9, 8)
The distances will be equal
∴ \sqrt {x^2-18x + 145} = \sqrt {x^2-6x + 13}
Squaring on both sides we get,
x^2 – 18x + 145 = x^2 – 6x + 13
-18x + 145 = -6x + 13
-18x + 6x = -145 + 13
-12x = -132
x = \frac {132}{12} = 11
The point is (11, 0)