| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular bisector of segment |
| Difficulty | Moderate -0.3 This is a standard coordinate geometry question testing routine skills: finding gradient, midpoint, perpendicular gradient, and using area of triangle. Part (c) requires slightly more thought (using perpendicular distance from line to find two positions), but all techniques are textbook exercises with no novel insight required. Slightly easier than average due to straightforward application of formulas. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m = \frac{2-11}{8+4}\) or \(m = \frac{11-2}{-4-8}\) | M1 | Correct gradient method, numerator and denominator both correct. Can also solve via simultaneous equations: \(11=-4m+c\), \(2=8m+c \Rightarrow 9=-12m\) |
| \(m = -\frac{3}{4}\) | A1 | Allow \(\frac{-3}{4}\) or \(\frac{3}{-4}\) or \(-0.75\). If equation of line found, must identify \(-\frac{3}{4}\) as gradient |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M\) is \(\left(2, \frac{13}{2}\right)\) | B1 | Allow unsimplified e.g. \(\left(\frac{-4+8}{2}, \frac{11+2}{2}\right)\). Condone lack of brackets |
| \(m_N = -1 \div -\frac{3}{4}\) | M1 | Applies perpendicular gradient rule to gradient from part (a) |
| \(y - \text{"}{\frac{13}{2}}\text{"} = \text{"}\frac{4}{3}\text{"}(x - \text{"}2\text{"})\) | M1 | Correct straight line method using midpoint and changed gradient. If using \(y=mx+c\), must reach \(c=\ldots\) |
| \(8x - 6y + 23 = 0\) | A1 | Allow any integer multiple |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB = \sqrt{(-4-8)^2+(11-2)^2}\ (=15)\) or \(AB^2=(-4-8)^2+(11-2)^2\ (=225)\) | M1 | Correct application of Pythagoras using points \(A\) and \(B\) (or \(\frac{1}{2}AB\) using \(M\)) |
| \(\frac{1}{2} \times MC \times AB = 37.5 \Rightarrow MC = \frac{75}{15}\ (=5)\) or \(MC^2=25\) | M1 | Uses \(AB\) and \(37.5\) correctly to find \(MC\) or \(MC^2\). May be implied by working to find \(AM\) or \(BM\) |
| \(m_N = \frac{4}{3}, MC=5 \Rightarrow C\) is \(\left(\text{"}2\text{"}-3,\ \frac{\text{"}13\text{"}}{2}-4\right)\) or \(\left(\text{"}2\text{"}+3,\ \frac{\text{"}13\text{"}}{2}+4\right)\) | dM1 | Correct strategy to find both \(x\) or both \(y\) coordinates. Dependent on both previous M marks. Condone slips in rearrangement |
| \((-1,\ 2.5)\) or \((5,\ 10.5)\) or \(x=-1, x=5\) or \(y=2.5, y=10.5\) | A1 | Does not have to be written as coordinates |
| \((-1,\ 2.5)\) and \((5,\ 10.5)\) | A1 | Both coordinates paired correctly. Condone lack of brackets |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m = \frac{2-11}{8+4}$ or $m = \frac{11-2}{-4-8}$ | M1 | Correct gradient method, numerator and denominator both correct. Can also solve via simultaneous equations: $11=-4m+c$, $2=8m+c \Rightarrow 9=-12m$ |
| $m = -\frac{3}{4}$ | A1 | Allow $\frac{-3}{4}$ or $\frac{3}{-4}$ or $-0.75$. If equation of line found, must identify $-\frac{3}{4}$ as gradient |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M$ is $\left(2, \frac{13}{2}\right)$ | B1 | Allow unsimplified e.g. $\left(\frac{-4+8}{2}, \frac{11+2}{2}\right)$. Condone lack of brackets |
| $m_N = -1 \div -\frac{3}{4}$ | M1 | Applies perpendicular gradient rule to gradient from part (a) |
| $y - \text{"}{\frac{13}{2}}\text{"} = \text{"}\frac{4}{3}\text{"}(x - \text{"}2\text{"})$ | M1 | Correct straight line method using midpoint and changed gradient. If using $y=mx+c$, must reach $c=\ldots$ |
| $8x - 6y + 23 = 0$ | A1 | Allow any integer multiple |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB = \sqrt{(-4-8)^2+(11-2)^2}\ (=15)$ or $AB^2=(-4-8)^2+(11-2)^2\ (=225)$ | M1 | Correct application of Pythagoras using points $A$ and $B$ (or $\frac{1}{2}AB$ using $M$) |
| $\frac{1}{2} \times MC \times AB = 37.5 \Rightarrow MC = \frac{75}{15}\ (=5)$ or $MC^2=25$ | M1 | Uses $AB$ and $37.5$ correctly to find $MC$ or $MC^2$. May be implied by working to find $AM$ or $BM$ |
| $m_N = \frac{4}{3}, MC=5 \Rightarrow C$ is $\left(\text{"}2\text{"}-3,\ \frac{\text{"}13\text{"}}{2}-4\right)$ or $\left(\text{"}2\text{"}+3,\ \frac{\text{"}13\text{"}}{2}+4\right)$ | dM1 | Correct strategy to find both $x$ or both $y$ coordinates. Dependent on both previous M marks. Condone slips in rearrangement |
| $(-1,\ 2.5)$ **or** $(5,\ 10.5)$ **or** $x=-1, x=5$ **or** $y=2.5, y=10.5$ | A1 | Does not have to be written as coordinates |
| $(-1,\ 2.5)$ **and** $(5,\ 10.5)$ | A1 | Both coordinates paired correctly. Condone lack of brackets |
---6. The point $A$ has coordinates $( - 4,11 )$ and the point $B$ has coordinates $( 8,2 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of the line $A B$, giving your answer as a fully simplified fraction.
The point $M$ is the midpoint of $A B$. The line $l$ passes through $M$ and is perpendicular to $A B$.
\item Find an equation for $l$, giving your answer in the form $p x + q y + r = 0$ where $p , q$ and $r$ are integers to be found.
The point $C$ lies on $l$ such that the area of triangle $A B C$ is 37.5 square units.
\item Find the two possible pairs of coordinates of point $C$.
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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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\hfill \mbox{\textit{Edexcel P1 2020 Q6 [11]}}