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Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2). - Mathematics

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प्रश्न

Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).

बेरीज
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उत्तर

Let the point on the y-axis be Y\[\left( 0, y, 0 \right)\]which is equidistant from the points P\[\left( 3, 1, 2 \right)\]and Q \[\left( 5, 5, 2 \right)\]

Then, PY = QY

Now,

`sqrt((3 - 0)^2 + (1 - y)^2 + (2 - 0)^2) = sqrt((5 - 0)^2 + (5 - y)^2 + (2 - 0)^2)`

`=> sqrt((3)^2 + (1 - y)^2 + (2)^2) = sqrt((5)^2 + (5 - y)^2 + (2)^2)`

`=> sqrt(9 + (1 - y)^2 + 4) = sqrt(25 + (5 - y)^2 + 4)`

`=> 9 + (1 - y)^2 + 4 = 25 + (5 - y)^2 + 4`

`=> 9 + (1 - y)^2 + cancel(4) = 25 + (5 - y)^2 + cancel(4)`

`=> 9 + (1 - y)^2 = 25 + (5 - y)^2`

`=> 1 + y^2 - 2y = 25 - 9 + (5 - y)^2`

`=> 1 + y^2 - 2y = 16 + 25 + y^2 - 10y`

`=> 1 + cancel(y^2) - 2y = 41 + cancel(y^2) - 10y`

`=> - 2y = 41 - 1 - 10y`

`=> - 2y = 40 - 10y`

⇒ 8y = 40
⇒ y = `40/8`

⇒ y = 5

Thus, the required point on the y-axis is (0, 5, 0).

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पाठ 28: Introduction to three dimensional coordinate geometry - Exercise 28.2 [पृष्ठ ९]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 28 Introduction to three dimensional coordinate geometry
Exercise 28.2 | Q 6 | पृष्ठ ९

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