Standard +0.3 This is a coordinate geometry problem requiring understanding of perpendicular bisectors and midpoints. Students must find the gradient of the perpendicular, use the midpoint lying on the given line, and solve simultaneous equations. While it involves multiple steps, these are all standard AS-level techniques with no novel insight required, making it slightly easier than average.
Point \(C\) has coordinates \((c, 2)\) and point \(D\) has coordinates \((6, d)\).
The line \(y + 4x = 11\) is the perpendicular bisector of \(CD\).
Find \(c\) and \(d\).
[5 marks]
Question 5:
5 | Forms an equation for gradient of
CD = ΒΌ or βΒΌ of the form
difference in y over difference in x
(or vice versa = 4 or -4) | AO3.1a | M1 | πβ2 1
=
6βπ 4
4d β 8 = 6 β c
c + 4d = 14
2+π π+6
+4( )= 11
2 2
4c + d = β4
c = β2 d = 4
Obtains a correct equation for c & d | AO1.1b | A1
Forms an equation for the mid-
point of CD lying on y + 4x = 11 | AO3.1a | M1
Obtains correct equation for c & d
(any correct form) | AO1.1b | A1
Solves for c and d CAO | AO1.1b | A1
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
Point $C$ has coordinates $(c, 2)$ and point $D$ has coordinates $(6, d)$.
The line $y + 4x = 11$ is the perpendicular bisector of $CD$.
Find $c$ and $d$.
[5 marks]
\hfill \mbox{\textit{AQA AS Paper 1 2018 Q5 [5]}}