Find the coordinates of a point on y-axis which are at a distance of $$5\sqrt{2}$$ from the point P (3, –2, 5).

Asked by Pragya Singh | 1 year ago |  208

##### Solution :-

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 $$5\sqrt{2}$$.

Now, by using distance formula,

We know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by

Distance of PQ =

$$\sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}$$

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

$$\sqrt{(3-0)^2 + (-2-y)^2 + (5-0)^2}$$

$$\sqrt{3^2 + (-2-y)^2 + 5^2}$$

$$\sqrt{(-2-y)^2 + 9 + 25}$$

$$5 \sqrt{2}$$ = $$\sqrt{(-2-y)^2 + 34}$$

Squaring on both the sides, we get

(-2 -y)2 + 34 = 25 × 2

(-2 -y)2 = 50 – 34

4 + y2 + (2 × -2 × -y) = 16

y2 + 4y + 4 -16 = 0

y2 + 4y – 12 = 0

y2 + 6y – 2y – 12 = 0

y (y + 6) – 2 (y + 6) = 0

(y + 6) (y – 2) = 0

y = -6, y = 2

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

Answered by Pragya Singh | 1 year ago

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