1.

What will be the point of y-axis which will be equidistant from the points (-1, 0) and (3, 9)?(a) (5, \( \frac {89}{18} \))(b) (1, \( \frac {89}{18} \))(c) (16, \( \frac {89}{18} \))(d) (0, \( \frac {89}{18} \))I got this question in an online quiz.My question is taken from Geometry topic in section Coordinate Geometry of Mathematics – Class 10

Answer»

Right OPTION is (d) (0, \( \FRAC {89}{18} \))

To EXPLAIN I would say: LET the point on y-axis be (0, y)

Distance between (-1, 0) and (0, y) = \( \SQRT {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(0 + 1)^2 + (y-0)^2} \)

= \( \sqrt {y^2 + (1)^2} \)

= \( \sqrt {y^2 + 1} \)

Distance between (3, 9) and (0, y) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(0-3)^2 + (y-9)^2} \)

= \( \sqrt {y^2-18y + 81 + (-3)^2} \)

= \( \sqrt {y^2-18y + 81 + 9} \)

= \( \sqrt {y^2-18y + 90} \)

Since, the point (0, y) is equidistant from (-1, 0) and (3, 9)

The distances will be equal

∴ \( \sqrt {y^2 + 1} =\sqrt {y^2-18y + 90} \)

Squaring on both sides we get,

y^2 + 1 = y^2 – 18y + 90

1 – 90 = -18y

-89 = -18y

y = \( \frac {89}{18} \)

The point is (0, \( \frac {89}{18} \))


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