Higher November 2021 Paper 1 Q17
17. The diagram shows the points \(P(-5, 16)\) and \(Q(5, -4)\), joined by a straight line.
Diagram not drawn to scale
M is the midpoint of PQ.
By finding the gradient of PQ and the coordinates of M, show that the equation of the perpendicular bisector of PQ is \(2y = x + 12\).
You must show all your working. [6]
| Scheme | Marks | Comments |
|---|---|---|
| \(-2\) | B2 | B1 for \(\dfrac{16 - -4}{-5 - 5}\) with at most one arithmetic error oe, si; may be implied by a right-angled triangle drawn with dimensions ‘their 20’ and \((-)10\) or 20 and ‘their \((-)10\)’ marked and ‘their \(\dfrac{y\ \text{diff}}{x\ \text{diff}}\)’, which must be negative, calculated; If no marks, award SC1 for a derived answer of 2. |
| (0, 6) | B2 | May be on diagram B1 for each correct coordinate or for sight of \(\left(\dfrac{-5 + 5}{2}, \dfrac{16 - 4}{2}\right)\) |
| \(y = \tfrac{1}{2}x + 6\) leading to \(2y = x + 12\) | B2 | FT ‘their 6’ and ‘their \(-2\)’ for B1 B1 for \(y = \tfrac{1}{2}x + c\) or for \(y = kx + 6\ \ k \ne 0\) If B2 B2 B0, award SC1 for showing \(2y = x + 12\) is the same as \(y = \tfrac{1}{2}x + 6\) and that \(-2 \times \dfrac{1}{2} = -1\) and that (0, 6) is on the line \(y = \tfrac{1}{2}x + 6\) |
| (6) |