Three points have coordinates \(A(0, 7)\), \(B(8, 3)\), and \(C(3k, k)\). Find the value of the constant \(k\) for which:
- \(C\) lies on the line that passes through \(A\) and \(B\).
- \(C\) lies on the perpendicular bisector of \(AB\).
Solution
Part (i):
The slope of line \(AB\) is \(m = \frac{3 - 7}{8 - 0} = -\frac{1}{2}\).
The equation of line \(AB\) is \(y = -\frac{1}{2}x + 7\).
Substitute \(C(3k, k)\) into the line equation:
\(k = -\frac{1}{2}(3k) + 7\)
\(k = -\frac{3}{2}k + 7\)
\(2k = -3k + 14\)
\(5k = 14\)
\(k = 2.8\)
Part (ii):
The midpoint of \(AB\) is \(M(4, 5)\).
The perpendicular slope is \(2\), so the equation of the perpendicular bisector is:
\(y - 5 = 2(x - 4)\)
\(y = 2x - 3\)
Substitute \(C(3k, k)\) into the bisector equation:
\(k = 2(3k) - 3\)
\(k = 6k - 3\)
\(5k = 3\)
\(k = 0.6\)
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