Given a line: $x-\frac{2}{2} = y-\frac{1}{6} = z+\frac{2}{2}$ and two points $P_1(1,1,0)$ and $P_2(0,1,-1)$, identify the point $V$ which resides on the line and is equidistant from $P_1$, and $P_2$.
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Hi, and welcome to MSE. Questions are received better if you explain what you've tried and where you are stuck. Assuming you're on a plane in euclidian space, there's a line that runs equidistant to any two points. It's orthogonal to the line between those points. Find where it intersects the other line. – it's a hire car baby Nov 24 '20 at 15:30
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The locus of all the point $P(x,y)$ such that $PP_1=PP_2$ is the plane $$(x-1)^2+(y-1)^2+z^2=x^2+(y-1)^2+(z+1)^2\to x+z=0$$ Find the intersection between the line and the plane solving the system $$\begin{cases} \frac{x-2}{2} =\frac{y-1}{6}\\ \frac{x-2}{2} = \frac{z+2}{2}\\ x+z=0\\ \end{cases} $$ Solution should be $V(0,-5,0)$
Raffaele
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