On a line

On a line p : 3 x - 4 y - 3 = 0, determine the point C equidistant from points A[4, 4] and B[7, 1].

Correct result:

x =  9
y =  6

Solution:

3x4y3=0  x=(3+4y)/3=9 y=(3x3)/4 AC=BC  (x4)2+(y4)2=(x7)2+(y1)2  (x4)2+((3x3)/44)2=(x7)2+((3x3)/41)2 1.5x=13.5 32x=272 3x=27 x=27/3=9 3x-4y-3=0 \ \\ \ \\ x=(3+4y)/3=9 \ \\ y=(3x-3)/4 \ \\ |AC|=|BC| \ \\ \ \\ (x-4)^2+(y-4)^2=(x-7)^2+(y-1)^2 \ \\ \ \\ (x-4)^2+((3x-3)/4-4)^2=(x-7)^2+((3x-3)/4-1)^2 \ \\ 1.5x=13.5 \ \\ \dfrac{ 3 }{ 2 }x=\dfrac{ 27 }{ 2 } \ \\ 3x=27 \ \\ x=27 / 3=9 \ \\
y=(3x3)/4=(3 93)/4=6y=(3x-3)/4=(3 \cdot \ 9-3)/4=6



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