The equation of line A is (x)^2 + 11x + 12 = y - 4, while the equation of line B is x - 6 = y + 2. Find the co-ordinate(s) of the point at which lines A and B intersect.
While this question may seem complicated, this question is simply asking you to solve the equations of these two lines as simultaneous equations. Line A: x2 + 11x + 12 = y - 4 --> x2 + 11x + 16 = y; Line B: x - 6 = y + 2 --> x - 8 = y. At the co-ordinate(s) at which lines A and B intersect, x2 + 11x + 16 = x - 8. If you bring all the x's in the equation above to the same side: x2 + 10x + 24 = 0, which can also be written as: (x + 6)(x + 4) = 0. Solving this equation for x: x + 6 = 0 (x = - 6) AND x + 4 = 0 (x = - 4)When x = - 6, y = (- 6) - 8 = - 14 AND when x = - 4, y = (- 4) - 8 = - 12... Therefore lines A and B cross at two points: (- 6, -14) and (-4, -12)
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