Find the coordinates of a point on Y-axis which are at a distance of 52 from the point P(3, – 2,5).

# 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).

1. A

(0 ,-6,0) or (0 ,2,0)

2. B

(0 ,6,0) or (0 ,-2,0)

3. C

(0 ,-6,0) or (1,2,0)

4. D

None of these

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### Solution:

Let any point on the Y-axis is A(0 ,y,0).
Given, distance between P and A, PA =$5\sqrt{2}$

On squaring both sides, we get

$\begin{array}{l}\left(3-0{\right)}^{2}+\left(-2-\mathrm{y}{\right)}^{2}+\left(5-0{\right)}^{2}=50\\ ⇒9+4+{\mathrm{y}}^{2}+4\mathrm{y}+25=50\\ ⇒{\mathrm{y}}^{2}+4\mathrm{y}+38-50=0\\ ⇒{\mathrm{y}}^{2}+4\mathrm{y}-12=0\\ ⇒\left(\mathrm{y}+6\right)\left(\mathrm{y}-2\right)=0\\ ⇒\mathrm{y}=-6,2\end{array}$

Hence, the points on Y-axis are (0 , - 6,0) or (0,2,0).

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