The curve y = 7 - \(\frac{6}{x}\) and the line y + 2x - 3 = 0 intersect at two point. Finf the; (a) coordinates of the two points (b) equation of the perpen...
Question 1 Report
The curve y = 7 - \(\frac{6}{x}\) and the line y + 2x - 3 = 0 intersect at two point. Finf the;
(a) coordinates of the two points
(b) equation of the perpendicular bisector of the line joining the two points
Answer Details
(a) The line \(y + 2x - 3 = 0\) gives \(y = 3 - 2x\). At intersection with \(y = 7 - \dfrac{6}{x}\):
\[7 - \frac{6}{x} = 3 - 2x\]Multiply through by \(x\):
\[7x - 6 = 3x - 2x^2 \;\Rightarrow\; 2x^2 + 4x - 6 = 0 \;\Rightarrow\; x^2 + 2x - 3 = 0\] \[(x+3)(x-1)=0 \;\Rightarrow\; x = -3 \text{ or } x = 1\]When \(x = 1:\ y = 3 - 2 = 1\). When \(x = -3:\ y = 3 + 6 = 9\).
The points are \((1,\ 1)\) and \((-3,\ 9)\).
(b) Midpoint \(= \left(\dfrac{1 + (-3)}{2},\ \dfrac{1 + 9}{2}\right) = (-1,\ 5)\).
Gradient of the join \(= \dfrac{9 - 1}{-3 - 1} = \dfrac{8}{-4} = -2\); the perpendicular gradient is \(\dfrac{1}{2}\).
\[y - 5 = \tfrac{1}{2}(x + 1) \;\Rightarrow\; 2y - 10 = x + 1 \;\Rightarrow\; x - 2y + 11 = 0\]Perpendicular bisector: \(x - 2y + 11 = 0\).
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