Question

Find the equation of the locus of a point whose distance from the xy-plane is equal to its distance
from the point (-1,2,-3) ​

Answer 1

Step-by-step explanation:

HOPE THIS ATTACHMENT HELPS

Answer 2

Answer:

The equation of the locus of a point whose distance from the XY-plane is equal to its distance from the point is given by ​$\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}+4 \mathrm{y}-6 \mathrm{z}+14=0$

Step-by-step explanation:

Let the point be $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$

So its distance from $\mathrm{xy}-$ the plane is $\mathrm{d}_{1}=|\mathrm{z}| (modulus of \mathrm{z} -coordinate )$

And its distance of this point from the point $(-1,2,-3)$ is $\mathrm{d}_{2}=$

$\sqrt{(\mathrm{x}-1)^{2}+(\mathrm{y}-2)^{2}+(\mathrm{z}-3)^{2}}$

Now, it is given that $\mathrm{d}_{1}=\mathrm{d}_{2} .$

$\begin{aligned}&\Rightarrow \mathrm{d}_{1}^{2}=\mathrm{d}_{2}^{2} \\&\Rightarrow(\mathrm{x}+1)^{2}+(\mathrm{y}-2)^{2}+(\mathrm{z}+3)^{2}=\mathrm{z}^{2} \\&\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}+4 \mathrm{y}-6 \mathrm{z}+14=0\\&\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}+4 \mathrm{y}-6 \mathrm{z}+14=0 \text { (expand above equation) }\end{aligned}$