Three points have coordinates \(A(2, 6)\), \(B(8, 10)\), and \(C(6, 0)\). The perpendicular bisector of \(AB\) meets the line \(BC\) at \(D\). Find:
- the equation of the perpendicular bisector of \(AB\) in the form \(ax + by = c\),
- the coordinates of \(D\).
Solution
To solve the problem, we follow these steps:
- Find the midpoint \(M\) of \(AB\):
\(M = \left( \frac{2 + 8}{2}, \frac{6 + 10}{2} \right) = (5, 8)\)
- Calculate the gradient of \(AB\):
\(\text{Gradient of } AB = \frac{10 - 6}{8 - 2} = \frac{2}{3}\)
- Find the gradient of the perpendicular bisector:
\(\text{Perpendicular gradient} = -\frac{3}{2}\)
- Use the point-slope form to find the equation of the perpendicular bisector:
\(y - 8 = -\frac{3}{2}(x - 5)\)
Expanding and rearranging gives:
\(2y + 3x = 31\)
- Find the equation of line \(BC\):
\(y = 5(x - 6) \Rightarrow y = 5x - 30\)
- Solve the system of equations:
\(2y + 3x = 31\)
\(y = 5x - 30\)
Substitute \(y = 5x - 30\) into \(2y + 3x = 31\):
\(2(5x - 30) + 3x = 31\)
\(10x - 60 + 3x = 31\)
\(13x = 91\)
\(x = 7\)
Substitute \(x = 7\) back into \(y = 5x - 30\):
\(y = 5(7) - 30 = 5\)
Thus, the coordinates of \(D\) are \((7, 5)\).
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