Question

What will be the point of x-axis which will be equidistant from the points (9, 8) and (3, 2)?

(a) (10, 0)

(b) (13, 0)

(c) (11, 0)

(d) (12, 0)

I had been asked this question during an interview.

My doubt stems from Geometry in section Coordinate Geometry of Mathematics – Class 10

Answer 1

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)