Higher November 2020 Paper 2R Q23
23 \(P\) and \(Q\) are two points.
The coordinates of \(P\) are (–1, 6)
The coordinates of \(Q\) are (5, –4)
Find an equation of the perpendicular bisector of \(PQ\).
Give your answer in the form \(ax + by + c = 0\) where \(a\), \(b\) and \(c\) are integers.
(6)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{-1 + 5}{2}, \dfrac{6 - 4}{2}\right)\) or \(\left(\dfrac{4}{2}, \dfrac{2}{2}\right)\) or (2, 1) | M1 |
| \(\dfrac{-4 - 6}{5 - -1}\) or \(\dfrac{6 - -4}{-1 - 5}\) or \(-\dfrac{10}{6}\) or \(-\dfrac{5}{3}\) | M1 |
| \(\dfrac{-1}{-\frac{10}{6}}\) or \(\dfrac{6}{10}\) or \(\dfrac{-1}{-\frac{5}{3}}\) or \(\dfrac{3}{5}\) or 0.6 | M1 |
\(1 = \dfrac{3}{5}(2) + c\) or \(c = -\dfrac{1}{5}\) or \(c = -\dfrac{2}{10}\) or \(c = -0.2\) | M1 |
| \(y = \dfrac{3}{5}x - \dfrac{1}{5}\) or \(y = 0.6x - 0.2\) or \(5y = 3x - 1\) | A1 |
| \(3x - 5y - 1 = 0\) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for finding midpoint
M1: indep for finding the gradient of \(PQ\)
M1: for finding the perpendicular gradient to \(PQ\)
(ft their stated gradient)
M1: dep on 1st and 3rd M1 for substituting ‘(2, 1)’ into \(y = \text{“}\tfrac{3}{5}\text{”}x + c\) or find the value of \(c\)
oe eg \(y - \text{“}1\text{”} = \text{“}\tfrac{3}{5}\text{”}(x - \text{“}2\text{”})\)
A1: for a correct equation in any form
A1: for \(3x - 5y - 1 = 0\) or
\(5y - 3x + 1 = 0\) or
\(6x - 10y - 2 = 0\) oe
accept in the form \(ax + by = -c\)
eg \(3x - 5y = 1\) or \(5y - 3x = -1\) oe
Alternative Mark Scheme for Q23
| Scheme | Marks |
|---|---|
| \((x + 1)^2 + (y - 6)^2\) or \((x - 5)^2 + (y + 4)^2\) | M1 |
| \((x + 1)^2 + (y - 6)^2 = (x - 5)^2 + (y + 4)^2\) | M1 |
| \(x^2 + 2x + 1 + y^2 - 12y + 36\) or \(x^2 - 10x + 25 + y^2 + 8y + 16\) | M1 |
| \(x^2 + 2x + 1 + y^2 - 12y + 36 = x^2 - 10x + 25 + y^2 + 8y + 16\) | M1 |
| eg \(2x + 1 - 12y + 36 = -10x + 25 + 8y + 16\) or \(12x + 37 = 20y + 41\) | A1 |
| \(3x - 5y - 1 = 0\) | A1 |
Notes
M1: using \(PA^2 = QA^2\) (for some point \(A\) on the line)
A1: for a correct linear equation in \(x\) and \(y\)
A1: for \(3x - 5y - 1 = 0\) oe