Find the coordinates of a point on y-axis which are at a distance of 5 -2 from the point P 3-2-5-

Let Q(0, y, 0) be any point on y axis. Then PQ=√((0 3)^2+(y+2)^2+(0 5)^2) = √(9+y^2+4+4y+25) = √(y^2+4y+38) But √(y^2+4y+38)=5√(2) Squaring both sides, we ha ...

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Standard X

Mathematics

Distance Formula

Find the coor...

Question

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

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Solution

Let $Q (0, y, 0)$ be any point on y-axis.

Then

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

= $\sqrt{9 + y^{2} + 4 + 4 y + 25}$

= $\sqrt{y^{2} + 4 y + 38}$

But $\sqrt{y^{2} + 4 y + 38} = 5 \sqrt{2}$

Squaring both sides, we have

$y^{2} + 4 y + 38 = 50 \Rightarrow y^{2} + 4 y - 12$

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

$\Rightarrow y = 2, - 6$

Thus coordinates of point Q are $(0, 2, 0)$ and $(0, - 6, 0) .$

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

Let Q(0, y, 0) be any point on y-axis. Then PQ=√(0−3)2+(y+2)2+(0−5)2 = √9+y2+4+4y+25 = √y2+4y+38 But √y2+4y+38=5√2 Squaring both sides, we have y2+4y+38=50 ⇒ y2+4y−12 =0 ⇒ (y−2)(y+6)=0 ⇒ y=2,−6 Thus coordinates of point Q are (0, 2, 0) and (0, −6, 0).

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