Q. 44.3( 26 Votes )

# Find the coordinates of a point on y-axis which are at a distance of 5√2 from the point P (3, –2, 5).

Answer :

Let the point on y-axis be A (0, y, 0).

Then, it is given that the distance between the points A (0, y, 0) and P (3, -2, 5) is .

Now, by Distance Formula, we know that the distance between two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) is given by .

So, the distance between the points A (0, y, 0) and P (3, -2, 5) is given by

Squaring both sides, we get

(-2 -y)^{2} + 34 = 25 × 2

⇒ (-2 -y)^{2} = 50 – 34

⇒ 4 + y^{2} + (2 × -2 × -y) = 16

⇒ y^{2} + 4y + 4 -16 = 0

⇒ y^{2} + 4y – 12 = 0

⇒ y^{2} + 6y – 2y – 12 = 0

⇒ y (y + 6) – 2 (y + 6) = 0

⇒ (y + 6) (y - 2) = 0

⇒ y = -6, y = 2

So, the points (0, 2, 0) and (0, -6, 0) are the required points on the y-axis.

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