Moderate -0.3 This is a straightforward perpendicular bisector question requiring standard techniques: finding the midpoint, calculating the gradient of AB, using the perpendicular gradient relationship (-1/m), and substituting into the given equation form. While it involves algebraic manipulation with parameters, the method is routine and commonly practiced. Slightly easier than average due to being a direct application of well-rehearsed coordinate geometry formulas.
8 Points \(A\) and \(B\) have coordinates \(( h , h )\) and \(( 4 h + 6,5 h )\) respectively. The equation of the perpendicular bisector of \(A B\) is \(3 x + 2 y = k\). Find the values of the constants \(h\) and \(k\).
Either \(\dfrac{5h-h}{4h+6-h} = \dfrac{2}{3}\) or \(-\dfrac{4h+6-h}{5h-h} = -\dfrac{3}{2}\)
*M1
Using \(m_1m_2 = -1\) appropriately to form an equation
OR
Gradient of bisector \(= -\dfrac{3}{2}\)
B1
Using gradient of \(AB\) and \(A\), \(B\) or midpoint \(\rightarrow \dfrac{2}{3}x + \dfrac{h}{3} = y\) oe
*M1
Obtain equation of \(AB\) using gradient from \(m_1m_2 = -1\) and a point
Substitute co-ordinates of one of the other points
*M1
Arrive at an equation in \(h\)
\(h = 2\)
A1
Midpoint is \(\left(\dfrac{5h+6}{2}, 3h\right)\) or \((8, 6)\)
B1FT
Algebraic expression or FT for numerical answer from 'their \(h\)'
Uses midpoint and 'their \(h\)' with \(3x + 2y = k\)
DM1
Substitutes 'their midpoint' into \(3x + 2y = k\). If \(y = -\dfrac{3}{2}x + c\) is used (expect \(c = 18\)) the method mark should be withheld until they \(\times 2\)
\(\rightarrow k = 36\) soi
A1
Total
7
## Question 8:
| Answer | Mark | Guidance |
|--------|------|----------|
| Gradient of bisector $= -\dfrac{3}{2}$ | B1 | |
| gradient $AB = \dfrac{5h-h}{4h+6-h}$ | *M1 | Attempt at $\dfrac{y-step}{x-step}$ |
| Either $\dfrac{5h-h}{4h+6-h} = \dfrac{2}{3}$ or $-\dfrac{4h+6-h}{5h-h} = -\dfrac{3}{2}$ | *M1 | Using $m_1m_2 = -1$ appropriately to form an equation |
| **OR** | | |
| Gradient of bisector $= -\dfrac{3}{2}$ | B1 | |
| Using gradient of $AB$ and $A$, $B$ or midpoint $\rightarrow \dfrac{2}{3}x + \dfrac{h}{3} = y$ oe | *M1 | Obtain equation of $AB$ using gradient from $m_1m_2 = -1$ and a point |
| Substitute co-ordinates of one of the other points | *M1 | Arrive at an equation in $h$ |
| $h = 2$ | A1 | |
| Midpoint is $\left(\dfrac{5h+6}{2}, 3h\right)$ or $(8, 6)$ | B1FT | Algebraic expression or FT for numerical answer from 'their $h$' |
| Uses midpoint and 'their $h$' with $3x + 2y = k$ | DM1 | Substitutes 'their midpoint' into $3x + 2y = k$. If $y = -\dfrac{3}{2}x + c$ is used (expect $c = 18$) the method mark should be withheld until they $\times 2$ |
| $\rightarrow k = 36$ soi | A1 | |
| **Total** | **7** | |
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8 Points $A$ and $B$ have coordinates $( h , h )$ and $( 4 h + 6,5 h )$ respectively. The equation of the perpendicular bisector of $A B$ is $3 x + 2 y = k$. Find the values of the constants $h$ and $k$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q8 [7]}}