Moderate -0.8 This is a straightforward coordinate geometry question requiring finding the midpoint of AB, calculating the gradient of AB, using the perpendicular gradient property, and substituting x=0 into the perpendicular bisector equation. All steps are routine applications of standard formulas with no problem-solving insight needed, making it easier than average but not trivial since it requires multiple coordinated steps.
2 Two points \(A\) and \(B\) have coordinates \(( 1,3 )\) and \(( 9 , - 1 )\) respectively. The perpendicular bisector of \(A B\) intersects the \(y\)-axis at the point \(C\). Find the coordinates of \(C\).
Can be seen in working, accept \(\left(\frac{10}{2}, \frac{2}{2}\right)\)
\(m_{AB} = -\frac{1}{2}\) oe
B1
\(C\) to \((5, 1)\) has gradient \(2\)
\*M1
Use of \(m_1 \times m_2 = -1\)
Forming equation of line \((y = 2x - 9)\)
DM1
Using their perpendicular gradient and their midpoint to form the equation.
\(C(0, -9)\) or \(y = -9\)
A1
Total: 5
**Question 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Midpoint of $AB$ is $(5, 1)$ | **B1** | Can be seen in working, accept $\left(\frac{10}{2}, \frac{2}{2}\right)$ |
| $m_{AB} = -\frac{1}{2}$ oe | **B1** | |
| $C$ to $(5, 1)$ has gradient $2$ | **\*M1** | Use of $m_1 \times m_2 = -1$ |
| Forming equation of line $(y = 2x - 9)$ | **DM1** | Using their perpendicular gradient and their midpoint to form the equation. |
| $C(0, -9)$ **or** $y = -9$ | **A1** | |
| **Total: 5** | | |
2 Two points $A$ and $B$ have coordinates $( 1,3 )$ and $( 9 , - 1 )$ respectively. The perpendicular bisector of $A B$ intersects the $y$-axis at the point $C$. Find the coordinates of $C$.\\
\hfill \mbox{\textit{CAIE P1 2019 Q2 [5]}}