Find the coordinates of a point on y-axis which are at a distance of
Question:

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

Solution:

If a point is on the y-axis, then x-coordinate and the z-coordinate of the point are zero.

Let $\mathrm{A}(0, b, 0)$ be the point on the $y$-axis at a distance of $5 \sqrt{2}$ from point $\mathrm{P}(3,-2,5)$. Accordingly, $\mathrm{AP}=5 \sqrt{2}$

$\therefore \mathrm{AP}^{2}=50$

$\Rightarrow(3-0)^{2}+(-2-b)^{2}+(5-0)^{2}=50$

$\Rightarrow 9+4+b^{2}+4 b+25=50$

$\Rightarrow b^{2}+4 b-12=0$

$\Rightarrow b^{2}+6 b-2 b-12=0$

$\Rightarrow(b+6)(b-2)=0$

$\Rightarrow b=-6$ or 2

Thus, the coordinates of the required points are (0, 2, 0) and (0, –6, 0).

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