The equation of a curve is \(xy = 12\) and the equation of a line \(l\) is \(2x + y = k\), where \(k\) is a constant.
In the case where \(k = 11\), find the coordinates of the points of intersection of \(l\) and the curve.
Solution
Given the equations:
1. Curve: \(xy = 12\)
2. Line: \(2x + y = 11\)
Substitute \(y = 11 - 2x\) from the line equation into the curve equation:
\(x(11 - 2x) = 12\)
\(11x - 2x^2 = 12\)
Rearrange to form a quadratic equation:
\(2x^2 - 11x + 12 = 0\)
Factor the quadratic equation:
\((2x - 3)(x - 4) = 0\)
Thus, \(x = \frac{3}{2}\) or \(x = 4\).
For \(x = \frac{3}{2}\), substitute back to find \(y\):
\(y = 11 - 2 \times \frac{3}{2} = 8\)
For \(x = 4\), substitute back to find \(y\):
\(y = 11 - 2 \times 4 = 3\)
Therefore, the points of intersection are \((1.5, 8)\) and \((4, 3)\).
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