Standard +0.3 This is a straightforward coordinate geometry problem requiring students to apply two standard conditions: the perpendicular bisector passes through the midpoint of AB, and its gradient is the negative reciprocal of AB's gradient. While it involves simultaneous equations and multiple steps, the techniques are routine and commonly practiced in P1.
6 Points \(A\) and \(B\) have coordinates \(( 8,3 )\) and \(( p , q )\) respectively. The equation of the perpendicular bisector of \(A B\) is \(y = - 2 x + 4\).
Find the values of \(p\) and \(q\).
Lines meet when \(-2x+4=\frac{1}{2}(x-8)+3\), solving as far as \(x=\)
*M1
Equating given perpendicular bisector with line through \((8,3)\) using *their* gradient of \(AB\) (but not \(-2\)) and solving. Expect \(x=2\), \(y=0\)
Using mid-point to get as far as \(p=\) or \(q=\)
DM1
Expect \(\frac{8+p}{2}=2\) or \(\frac{3+q}{2}=0\)
\(p=-4,\ q=-3\)
A1
Allow coordinates of \(B\) are \((-4,-3)\)
Alternative method:
Gradient \(AB=\frac{1}{2}\)
B1
SOI
\(\frac{q-3}{p-8}=\frac{1}{2}\) [leading to \(2q=p-2\)], \(\frac{q+3}{2}=-2\left(\frac{8+p}{2}\right)+4\) [leading to \(q=-11-2p\)]
*M1
Equating gradient of \(AB\) with *their* gradient of \(AB\) (but not \(-2\)) and using mid-point in equation of perpendicular bisector
Solving simultaneously *their* 2 linear equations
DM1
Equating and solving 2 correct equations as far as \(p=\) or \(q=\)
\(p=-4,\ q=-3\)
A1
Allow coordinates of \(B\) are \((-4,-3)\)
4
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient $AB=\frac{1}{2}$ | B1 | SOI |
| Lines meet when $-2x+4=\frac{1}{2}(x-8)+3$, solving as far as $x=$ | *M1 | Equating given perpendicular bisector with line through $(8,3)$ using *their* gradient of $AB$ (but not $-2$) and solving. Expect $x=2$, $y=0$ |
| Using mid-point to get as far as $p=$ or $q=$ | DM1 | Expect $\frac{8+p}{2}=2$ or $\frac{3+q}{2}=0$ |
| $p=-4,\ q=-3$ | A1 | Allow coordinates of $B$ are $(-4,-3)$ |
| **Alternative method:** | | |
| Gradient $AB=\frac{1}{2}$ | B1 | SOI |
| $\frac{q-3}{p-8}=\frac{1}{2}$ [leading to $2q=p-2$], $\frac{q+3}{2}=-2\left(\frac{8+p}{2}\right)+4$ [leading to $q=-11-2p$] | *M1 | Equating gradient of $AB$ with *their* gradient of $AB$ (but not $-2$) and using mid-point in equation of perpendicular bisector |
| Solving simultaneously *their* 2 linear equations | DM1 | Equating and solving 2 correct equations as far as $p=$ or $q=$ |
| $p=-4,\ q=-3$ | A1 | Allow coordinates of $B$ are $(-4,-3)$ |
| | **4** | |
6 Points $A$ and $B$ have coordinates $( 8,3 )$ and $( p , q )$ respectively. The equation of the perpendicular bisector of $A B$ is $y = - 2 x + 4$.
Find the values of $p$ and $q$.\\
\hfill \mbox{\textit{CAIE P1 2021 Q6 [4]}}