Mathematics · May / June 2024 · Paper 11 · Question 9
YMA01
As printed in the paper
The question as text
9. Given that • the point A has coordinates (4, 2) • the point B has coordinates (15, 7) • the line l1 passes through A and B
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
9(a)[3]
7
2
5
−
m
=
=
15
−
4
11
5
5
y
2
(
x
4
)
or
y
7
(
x
15
)
−
=
−
−
=
−
11
11
5
5
2
or
y
=
x
+
c
2
=
×+
4
c
c
=
...
11
11
11
5
x
−
11
y
+
2
=
0
(a) ALT
2
4
m
c
=
+
y
=
mx
+
c
7
15
m
c
=
+
2
=
4
m
+
c
5
2
m
=
...
,
c
=
...
7
=
15
m
+
c
11
11
5
x
−
11
y
+
2
=
0
9(b)[4]
(
)
2
(
)
2
(
)
2
15
−
x
+
7
−
2
=
5 5
B
or e.g.
(
15
x
)
2
(
7
2
)
2
5 5
−
+
−
=
A
or e.g.
C
D
C
(
)
2
5 5
=
5
2
+
CD
2
oe
Way 1: (
)
2
(
)
2
15
−
x
+
25
=
125
15
−
x
=
100
x
=
...
or e.g.
Way 2:
125
=
25
+
CD
2
CD
2
=
100
CD
=
10
x
=
...
(
5, 2 or (
)
)
25, 2
(
5, 2 and (
)
)
25, 2
9(c)[2]
1
(
)
Area
=
"5" 4
−
×
5
2
5
=
2
Notes:
(a)
M1: Correct method for the gradient
M1: For a correct straight line method using their gradient and point A or point B with x and y correctly placed.
Alternatively uses y = mx + c and reaches c = …
y
−
2
x
−
4
Note that
=
scores both M marks.
7
−
2
15
−
4
A1: Correct equation in the required form including the “= 0”. Allow any integer multiple of this equation.
ALT
M1: Substitutes x = 4, y = 2 and x = 15, y = 7 into the general equation y = mx + c to obtain 2 equations
in m and c.
M1: Solves 2 equations of the form
α
=
m
β
+ to find values for m and c.
c
A1: Correct equation in the required form including the “= 0”. Allow any integer multiple of this equation.
B(15, 7)
5 5
A(4, 2)
C
D
C
(b)
M1: Correct application of Pythagoras for triangle BCD (see diagram above)
5 5 for (
2
The statements shown are sufficient for this mark but condone missing brackets e.g.
long as the intention is clear.
M1: A complete, correct and full method to obtain at least one x coordinate for C.
(
)
2
For Way 1 this mark is for processing an equation of the form: (
)
2
15
−
x
+
α
2
=
5 5
where
0
<
α
2
<
125
to obtain at least one value for x.
For Way 2 this mark is for processing an equation of the form: (
)
2
5 5
=
CD
2
+
α
2
where
0
<
to obtain at least one value for CD and then attempts at least one value of x either 15 – CD or 15 + CD.
Condone the use of x (or any other letter) for CD.
The work for this mark must be correct so e.g. (
)
(
)
(
) (
)
15
−
x
2
+
7
−
2
2
=
5 5
15
−
x
+
7
−
2
or e.g. (
)
2
5 5
=
CD
2
+
α
2
5 5
=
CD
+
α
scores M0. If a 3TQ is being solved, the usual rules apply.
A1: At least one correct position for C.
May be seen as a coordinate pair with or without brackets or as e.g. x = 5, y = 2
A1: Both correct pairs of coordinates.
May be seen as coordinate pairs with or without brackets or as e.g. x = 5, y = 2
Note that some candidates think the equation for l2 is y = 4 rather than y = 2
This scores a maximum of 1 mark in part (b) and will look like this:
(
)
2
(
)
2
(
)
2
(
)
2
(
)
2
15
−
x
+
7
−
4
=
5 5
15
−
x
+
7
−
4
=
125
−
9
=
116
(
15
)
2 29
15
2 29
−
x
= ±
x
=
±
or
(
)
2
5 5
3
2
2
2
116
2 29
=
+
CD
CD
=
CD
=
x
=
15
±
2 29
This scores M0M1A0A0
(c)
M1: Applies a correct triangle area for one of their points of the form (k, 2), k ≠ 15. May be implied.
1
(
)
This mark is for e.g.
2 k −
4
× where k is one of their calculated x coordinates from part (b)
5
1 11
(
)
or for
±
CD
× where CD is their calculated value from part (b)
5
2
1
4
15
"5"
4
1 4 7 15 2
May see shoelace method:
=
×+
×+
"5" 2
×−×−×
2 4
7 "5" 2 15
−×
2
2
7
2
2
2
Allow other correct methods for the area e.g. using trigonometry.
Their C must be of the form (k, 2) and a correct formula used or implied.
A1: Correct minimum area. Must be 2.5 not awrt 2.5 (e.g. if they use trigonometry and get an inexact answer)
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