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June 2013 p11 q7
102
A curve is given by the equation \(y = x^2 - 4x + 4\) and a line by the equation \(y = mx\), where \(m\) is a constant. For \(m = 1\), the curve and the line intersect at points \(A\) and \(B\). Find the coordinates of the midpoint of \(AB\).
Solution
To find the points of intersection, set the equations equal: \(x^2 - 4x + 4 = x\).
Rearrange to form a quadratic equation: \(x^2 - 5x + 4 = 0\).
Factor the quadratic: \((x - 1)(x - 4) = 0\).
Thus, \(x = 1\) and \(x = 4\).
Substitute back to find \(y\) coordinates: \(y = 1\) when \(x = 1\) and \(y = 4\) when \(x = 4\).
Points \(A\) and \(B\) are \((1, 1)\) and \((4, 4)\).
The midpoint of \(AB\) is \(\left( \frac{1 + 4}{2}, \frac{1 + 4}{2} \right) = \left( \frac{5}{2}, \frac{5}{2} \right)\).