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].

Result

x =  2
y =  -2

Solution:

x=4+t y=3+2t  (4+t−1)2+(3+2⋅ t−2)2=(4+t−(−1))2+(3+2⋅ t−0)2  12t=−24  12t=−24  t=−2  x=4+t=4+(−2)=2x=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 \ \\ \ \\ 12t=-24 \ \\ \ \\ t=-2 \ \\ \ \\ x=4+t=4+(-2)=2
y=3+2⋅ t=3+2⋅ (−2)=−1=−2y=3+2 \cdot \ t=3+2 \cdot \ (-2)=-1=-2

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