| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Perpendicular bisector of chord |
| Difficulty | Moderate -0.5 This is a multi-part question covering standard C1 coordinate geometry: finding line equations, perpendicular bisectors, and circle properties. While it has multiple steps (4 parts), each individual part uses routine techniques (gradient formula, midpoint, perpendicular gradient, substitution). The 'show that' parts reduce difficulty by providing target answers. Slightly easier than average due to straightforward application of formulas with no novel problem-solving required. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| grad perp \(= -1/\text{grad AB}\) stated, or used after their grad AB stated in this part | M1 | or showing \(3 \times -\frac{1}{3} = -1\); if (i) is wrong, allow first M1 here ft, provided answer is correct ft |
| midpoint \([\text{of AB}] = (2, 2)\) | M1 | must state 'midpoint' or show working |
| \(y - 2 = \text{their grad perp } (x - 2)\) or ft their midpoint | M1 | for M3 this must be correct, starting from grad \(AB = -\frac{1}{3}\), and also needs correct completion to give ans \(y = 3x - 4\) |
| Alt method working back from ans: | mark one method or the other, to benefit of candidate, not a mixture | |
| grad perp \(= -1/\text{grad AB}\) and showing/stating same as given line | M1 | e.g. stating \(-\frac{1}{3} \times 3 = -1\) |
| finding intersection of their \(y = -\frac{1}{3}x - \frac{8}{3}\) and \(y = 3x - 4\) is \((2, 2)\) | M1 | or showing that \((2, 2)\) is on \(y = 3x - 4\), having found \((2, 2)\) first |
| showing midpoint of AB is \((2, 2)\) | M1 | [for both methods: for M3 must be fully correct] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| subst \(x = 3\) into \(y = 3x - 4\) and obtaining centre \(= (3, 5)\) | M1 | or using \((-1-3)^2 + (3-b)^2 = (5-3)^2 + (1-b)^2\) and finding \((3, 5)\) |
| \(r^2 = (5-3)^2 + (1-5)^2\) o.e. | M1 | or \((-1-3)^2 + (3-5)^2\) or ft their centre using A or B |
| \(r = \sqrt{20}\) o.e. cao | A1 | |
| equation is \((x-3)^2 + (y-5)^2 = 20\) or ft their \(r\) and \(y\)-coord of centre | B1 | condone \((x-3)^2 + (y-b)^2 = r^2\) o.e. or \((x-3)^2 + (y - \text{their } 5)^2 = r^2\) o.e. (may be seen earlier) |
## Question 1(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| grad perp $= -1/\text{grad AB}$ stated, or used after their grad AB stated in this part | M1 | or showing $3 \times -\frac{1}{3} = -1$; if (i) is wrong, allow first M1 here ft, provided answer is correct ft |
| midpoint $[\text{of AB}] = (2, 2)$ | M1 | must state 'midpoint' or show working |
| $y - 2 = \text{their grad perp } (x - 2)$ or ft their midpoint | M1 | for M3 this must be correct, starting from grad $AB = -\frac{1}{3}$, and also needs correct completion to give ans $y = 3x - 4$ |
| **Alt method working back from ans:** | | mark one method or the other, to benefit of candidate, not a mixture |
| grad perp $= -1/\text{grad AB}$ and showing/stating same as given line | M1 | e.g. stating $-\frac{1}{3} \times 3 = -1$ |
| finding intersection of their $y = -\frac{1}{3}x - \frac{8}{3}$ and $y = 3x - 4$ is $(2, 2)$ | M1 | or showing that $(2, 2)$ is on $y = 3x - 4$, having found $(2, 2)$ first |
| showing midpoint of AB is $(2, 2)$ | M1 | [for both methods: for M3 must be fully correct] |
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## Question 1(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| subst $x = 3$ into $y = 3x - 4$ and obtaining centre $= (3, 5)$ | M1 | or using $(-1-3)^2 + (3-b)^2 = (5-3)^2 + (1-b)^2$ and finding $(3, 5)$ |
| $r^2 = (5-3)^2 + (1-5)^2$ o.e. | M1 | or $(-1-3)^2 + (3-5)^2$ or ft their centre using A or B |
| $r = \sqrt{20}$ o.e. cao | A1 | |
| equation is $(x-3)^2 + (y-5)^2 = 20$ or ft their $r$ and $y$-coord of centre | B1 | condone $(x-3)^2 + (y-b)^2 = r^2$ o.e. or $(x-3)^2 + (y - \text{their } 5)^2 = r^2$ o.e. (may be seen earlier) |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{01569a16-66ba-422e-a74d-6f9430dd245b-1_520_1122_357_551}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
Fig. 11 shows the line through the points $\mathrm { A } ( - 1,3 )$ and $\mathrm { B } ( 5,1 )$.\\
(i) Find the equation of the line through $\mathbf { A }$ and $\mathbf { B }$.\\
(ii) Show that the area of the triangle bounded by the axes and the line through A and B is $\frac { 32 } { 3 }$ square units.\\
(iii) Show that the equation of the perpendicular bisector of AB is $y = 3 x - 4$.\\
(iv) A circle passing through A and B has its centre on the line $x = 3$. Find the centre of the circle and hence find the radius and equation of the circle.
\hfill \mbox{\textit{OCR MEI C1 Q1 [12]}}