| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Moderate -0.3 This is a straightforward P1 coordinate geometry question requiring standard techniques: finding perpendicular gradient (negative reciprocal), writing line equation through origin, finding intersection point, and calculating triangle area using coordinates. All steps are routine textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a) | 2x + 3y = 26 (cid:159)3y = 26 ± 2x and attempt to find m from y = mx + c | M1 |
| Answer | Marks |
|---|---|
| 3 3 3 | A1 |
| Answer | Marks |
|---|---|
| theirgradient 2 | M1 |
| Answer | Marks |
|---|---|
| 2 | A1 |
| Answer | Marks |
|---|---|
| (b) | 3 |
| Answer | Marks |
|---|---|
| 2 | M1 |
| Solves their equation in x or in y to obtain x = or y = | dM1 |
| Answer | Marks |
|---|---|
| 39 | A1 |
| Answer | Marks |
|---|---|
| 3 | B1 |
| Answer | Marks |
|---|---|
| 3 | 1 "26" |
| Answer | Marks |
|---|---|
| 2 3 | dM1 |
| Answer | Marks |
|---|---|
| and denominator) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
Question 8:
--- 8(a) ---
8(a) | 2x + 3y = 26 (cid:159)3y = 26 ± 2x and attempt to find m from y = mx + c | M1
26 2 2
( (cid:159) y (cid:32) (cid:16) x ) so gradient = (cid:16)
3 3 3 | A1
(cid:16)1 3
Gradient of perpendicular = (= )
theirgradient 2 | M1
3
Line goes through (0, 0) so y (cid:32) x
2 | A1
(4)
(b) | 3
Solves their y (cid:32) x with their 2x + 3y = 26 to form equation in x or in y
2 | M1
Solves their equation in x or in y to obtain x = or y = | dM1
156
x = 4 or any equivalent e.g. or y = 6 o.a.e
39 | A1
26
B= (0, ) used or stated in (b)
3 | B1
26
x=4
3 | 1 "26"
Area = (cid:117)"4"(cid:117)
2 3 | dM1
52
(cid:32) (o.e. with integer numerator
3
and denominator) | A1
(6)
(10 marks)
Notes:
(a)
Complete method for finding gradient. (This may be implied by later correct answers.) e.g.
M1:
Rearranges 2x(cid:14)3y(cid:32)26(cid:159) y(cid:32)mx(cid:14)c so m =
Or finds coordinates of two points on line and finds gradient e.g.
8(cid:16)0
(13,0) and (1,8) so m(cid:32)
1(cid:16)13
2 2
(cid:32)(cid:16) (cid:32)(cid:16) x
A1: States or implies that gradient condone if they continue correctly. Ignore errors
3 3
in constant term in straight line equation.
(cid:16)1
M1: Uses m ×m = –1 to find the gradient of l . This can be implied by the use of
1 2 2 theirgradient
3 3 39
A1: y (cid:32) x or 2y – 3x = 0 Allow y (cid:32) x(cid:14)0 Also accept 2y = 3x, y= x or even
2 2 26
3
y(cid:16)0(cid:32) (x(cid:16)0)and isw.
2
Question 8 notes continued
(b)
3
M1: Eliminates variable between their y (cid:32) x and their (possibly rearranged) 2x(cid:14)3y(cid:32)26 to
2
form an equation in x or y. (They may have made errors in their rearrangement).
dM1: (Depends on previous M mark) Attempts to solve their equation to find the value of x or y
A1: x = 4 or equivalent or y = 6 or equivalent
26 26
B1: y coordinate of B is (stated or implied) - isw if written as ( , 0).
3 3
Must be used or stated in (b)
dM1: (Depends on previous M mark) Complete method to find area of triangle OBC (using their
26
values of x and/or y at point C and their )
3
52 104 1352
A1: Cao or or o.e
3 6 78
Alternative 1
Uses the area of a triangle formula ½×OB ×(x coordinate of C)
Alternative methods: Several Methods are shown below. The only mark which differs from
Alternative 1 is the last M mark and its use in each case is described below:
Alternative 2
1
In 8(b) using (cid:117)BC(cid:117)OC
2
4
dM1: Uses the area of a triangle formula ½×BC ×OC (cid:36)(cid:79)(cid:86)(cid:82) (cid:73)(cid:76)(cid:81)(cid:71)(cid:86) (cid:50)(cid:38) (cid:11)(cid:32)(cid:165)52 (cid:12) (cid:68)(cid:81)(cid:71) (cid:37)(cid:38)(cid:32) (cid:11) 13 )
3
Alternative 3
1 0 4 0 0
In 8(b) using
2 0 6 260
3
1 0 4 0 0
dM1: States the area of a triangle formula or equivalent with their values
2 0 6 260
3
Alternative 4
In 8(b) using area of triangle OBX – area of triangle OCX where X is point (13, 0)
1 26 1
dM1: Uses the correct subtraction (cid:117)13(cid:117)" "(cid:16) (cid:117)13(cid:117)"6"
2 3 2
Alternative 5
In 8(b) using area = ½ (6 × 4) + ½ (4 × 8/3) drawing a line from C parallel to the x axis and
dividing triangle into two right angled triangles
dM1: For correct method (cid:68)(cid:85)(cid:72)(cid:68) (cid:32) (cid:242) (cid:11)(cid:179)(cid:25)(cid:180) × (cid:179) (cid:23)(cid:180)(cid:12) (cid:14) (cid:242) (cid:11)(cid:179)(cid:23)(cid:180) × (cid:62)(cid:179)2(cid:25)(cid:18)3(cid:180)-(cid:179)(cid:25)(cid:180)(cid:64)(cid:12)
Method 6 Uses calculus
4 26 2x 3x (cid:170)26 x2 3x2(cid:186) 4
dM1: (cid:179)" "(cid:16) (cid:16) dx= x(cid:16) (cid:16)
(cid:171) (cid:187)
3 3 2 (cid:172) 3 3 4 (cid:188)
0 0
Question | Scheme | Marks\includegraphics{figure_2}
The line $l_1$ shown in Figure 2 has equation $2x + 3y = 26$
The line $l_2$ passes through the origin $O$ and is perpendicular to $l_1$
\begin{enumerate}[label=(\alph*)]
\item Find an equation for the line $l_2$ [4]
\end{enumerate}
The line $l_1$ intersects the line $l_1$ at the point $C$. Line $l_1$ crosses the $y$-axis at the point $B$ as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the area of triangle $OBC$. Give your answer in the form $\frac{a}{b}$, where $a$ and $b$ are integers to be found. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2018 Q8 [10]}}