Standard +0.3 This is a straightforward coordinate geometry question requiring finding the perpendicular bisector equation (midpoint and perpendicular gradient) then solving simultaneously with a given line. All steps are standard AS-level techniques with no novel insight required, making it slightly easier than average.
8 The point A has coordinates \(( - 1 , - 2 )\) and the point B has coordinates (7,4). The perpendicular bisector of \(A B\) intersects the line \(y + 2 x = k\) at \(P\).
Determine the coordinates of P in terms of \(k\).
Gradient of AB is \(\frac{4-(-2)}{7-(-1)} = \frac{6}{8} = \frac{3}{4}\)
M1
Must be correct way up – allow one slip
Gradient of perpendicular \(-\frac{4}{3}\)
M1
Negative reciprocal FT their gradient of AB
Equation of perpendicular bisector \(y - 1 = -\frac{4}{3}(x-3)\)
M1
FT their gradient. Condone A or B used instead of M. Do not allow if their gradient of AB used
Solve simultaneous equations \(4x + 3y = 15\) and \(2x + y = k\); \(y = 15-2k\), \(x = \frac{1}{2}(3k-15)\)
M1, A1
Attempt to eliminate one variable; cao
P is \(\left(\frac{1}{2}(3k-15),\ 15-2k\right)\)
A1 [7]
cao
OR method:
Answer
Marks
Guidance
Answer
Marks
Guidance
P is of the form \((x, k-2x)\)
B1
Seen or implied
\((x+1)^2 + (y+2)^2 = (x-7)^2 + (y-4)^2\)
M1
Finding at least one distance
Equating distances
M1
Equating distances
\((x+1)^2 + (k-2x+2)^2 = (x-7)^2 + (k-2x-4)^2\)
M1
Substituting for \(y\)
\(-8x + 1 = 60 - 12k\)
M1
Attempt to simplify
\(x = \frac{1}{2}(3k-15)\) and \(y = 15-2k\)
A1
cao
P is \(\left(\frac{1}{2}(3k-15),\ 15-2k\right)\)
A1 [7]
cao
## Question 8:
**EITHER method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Midpoint AB is $(3, 1)$ | B1 | Seen or implied |
| Gradient of AB is $\frac{4-(-2)}{7-(-1)} = \frac{6}{8} = \frac{3}{4}$ | M1 | Must be correct way up – allow one slip |
| Gradient of perpendicular $-\frac{4}{3}$ | M1 | Negative reciprocal FT their gradient of AB |
| Equation of perpendicular bisector $y - 1 = -\frac{4}{3}(x-3)$ | M1 | FT their gradient. Condone A or B used instead of M. Do not allow if their gradient of AB used |
| Solve simultaneous equations $4x + 3y = 15$ and $2x + y = k$; $y = 15-2k$, $x = \frac{1}{2}(3k-15)$ | M1, A1 | Attempt to eliminate one variable; cao |
| P is $\left(\frac{1}{2}(3k-15),\ 15-2k\right)$ | A1 [7] | cao |
**OR method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| P is of the form $(x, k-2x)$ | B1 | Seen or implied |
| $(x+1)^2 + (y+2)^2 = (x-7)^2 + (y-4)^2$ | M1 | Finding at least one distance |
| Equating distances | M1 | Equating distances |
| $(x+1)^2 + (k-2x+2)^2 = (x-7)^2 + (k-2x-4)^2$ | M1 | Substituting for $y$ |
| $-8x + 1 = 60 - 12k$ | M1 | Attempt to simplify |
| $x = \frac{1}{2}(3k-15)$ and $y = 15-2k$ | A1 | cao |
| P is $\left(\frac{1}{2}(3k-15),\ 15-2k\right)$ | A1 [7] | cao |
---
8 The point A has coordinates $( - 1 , - 2 )$ and the point B has coordinates (7,4). The perpendicular bisector of $A B$ intersects the line $y + 2 x = k$ at $P$.
Determine the coordinates of P in terms of $k$.
\hfill \mbox{\textit{OCR MEI AS Paper 1 2020 Q8 [7]}}