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If: 3x-y = 5 and 2x2+y2 = 129 Then: 3x-y = 5 => y = 3x-5 And so: 2x2+(3x-5)2 = 129 => 11x2-30x-104 = 0 Using the quadratic equation formula: x = 52/11 and x = -2 By substitution points of intersection are: (52/11, 101/11) and (-2, -11)
2x2 + 3x - 20 = (x + 4)(2x - 5).
The locus of points equidistant from lines y = 0 and x = 3 is the line y = -x + 3.