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 (x1, y1, z1) and Q (x2, y2, z2) 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 + 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
So, the points (0, 2, 0) and (0, -6, 0) are the required points on the y-axis.
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