Question

What will be the value of x, if the distance between the points (5, 11) and (2, x) is 10?

(a) -11 + √91, -11 – √91

(b) 11 + √91, 11 – √91

(c) 11 + √91, 11 + √91

(d) -11 + √91, 11 – √91

The question was asked in a national level competition.

This question is from Geometry in division Coordinate Geometry of Mathematics – Class 10

Answer 1

Right option is (b) 11 + √91, 11 – √91

The explanation is: Distance between (5, 11) and (2, x) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(2-5)^2 + (x-11)^2} \)

= \( \sqrt {x^2-22x + 121 + (-3)^2} \)

= \( \sqrt {x^2-22x + 121 + 9} \)

= \( \sqrt {x^2-22x + 130} \)

The distance between (5, 11) and (2, x) is 10

∴ \( \sqrt {x^2-22x + 130} \) = 10

Squaring on both sides we get,

x^2 – 22x + 130 = 100

x^2 – 22x + 130 – 100 = 0

x^2 – 22x + 30 = 0

x = 11 + √91, 11 – √91