Find the locus of a point whose distance from x-axis is twice the distance from the point (1,-1,2):

Question

Find the locus of a point whose distance from $x$ -axis is twice the distance from the point $(1, - 1, 2)$ :

Smart Achievers · Accepted Answer

(3)

$Let the point P (x_{p}, y_{p}, z_{p}) be the required point,$

$the distance of the point from x-axis is \sqrt{y_{p}^{2} + z_{p}^{2}}, The distance from the point (1, - 1, 2) is$

$\Rightarrow$ $y_{p}^{2} + z_{p}^{2} = {(x_{p} - 1)}^{2} + {(y_{p} + 1)}^{2} + {(z_{p} - 2)}^{2} \Rightarrow x_{p}^{2} - 2 x_{p} + 2 y_{p} - 4 z_{p} + 6 = 0$

$Therefore, the locus of point P is$ $x^{2} - 2 x + 2 y - 4 z + 6 = 0$

Solution

Q.
Find the locus of a point whose distance from $x$ -axis is twice the distance from the point $(1, - 1, 2)$ :

1 $y^{2} + 2 x - 2 y - 4 z + 6 = 0$

2 $x^{2} + 2 x - 2 y - 4 z + 6 = 0$

3 $x^{2} - 2 x + 2 y - 4 z + 6 = 0$

4 $z^{2} - 2 x + 2 y - 4 z + 6 = 0$

Want Us to Email you About Special Offers & Updates?

Solution

Q. Find the locus of a point whose distance from x-axis is twice the distance from the point (1,-1,2):

1 y2+2x-2y-4z+6=0

2 x2+2x-2y-4z+6=0

3 x2-2x+2y-4z+6=0

4 z2-2x+2y-4z+6=0