On line

On line p: x = 4 + t, y = 3 + 2t, t is R, find point C, which has the same distance from points A [1,2] and B [-1,0].

Correct result:

x =  2
y =  -2

Solution:

$$\begin{gathered} x=4+t \\ y=3+2t \\ (4+t-1{)}^2+(3+2\cdot t-2{)}^2=(4+t-(-1){)}^2+(3+2\cdot t-0{)}^2 \\ 12t=-24 \\ 12\cdot t=-24 \\ 12t=-24 \\ t=-2 \\ x=4+t=4+(-2)=2 \end{gathered}$$

$y=3+2\cdot t=3+2\cdot (-2)=-1=-2$

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