Mathematics · Oct / Nov 2021 · Paper 11 · Question 5
YMA01
As printed in the paper
The question as text
5. The line l1 has equation 3y – 2x = 30 The line l2 passes through the point A(24, 0) and is perpendicular to l1 Lines l1 and l2 meet at the point P
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[5]
2
3
y
2
x
30
m =
−
=
⇒
3
3
3
y
−
0
3
y
= −
(
x
−
24),
y
= −
x
+
36, 2
y
+
3
x
=
72,
= −
2
2
x
−
24
2
2
3
(
)
Full method to find one coordinate of P E.g. Solves
x
+
10
= −
x
−
24
3
2
Coordinates of P (
)
12,18
5(b)[3]
y
=
0, 3
y
−
2
x
=
30
⇒
x
=
...
1
(
)
Area BPA =
2 ×
24
+
''15''
×
''18''
=
351
(a)
Gradient of 1l = 2
B1
3 seen or implied
Full method for equation of 2l . This involves an equation using the point (
M1
24,0
with a gradient using the negative reciprocal of their 2
3 . If they use y = mx + c then
must reach as far as c = …
3 (
A1ft
Correct equation of normal in any form e.g.
y
= −
x
−
24)
but condone poor
2
3 (
l
= −
x
−
24)
notation here and allow e.g. 2
provided this is used appropriately to
2
solve simultaneously. Follow through their negative reciprocal gradient.
M1
Full method to find one coordinate of P
Note that we allow a calculator to be used here e.g.
3
y
−
2
x
=
30, 2
y
+
3
x
=
72
⇒
x
=
... or
y
=
...
Coordinates of P (
)
12, 18 or x = 12, y = 18
A1
Alt (a)
Note that the first three marks can be scored via
Attempts 2l via 2
y
+
3
x
=
c
B1
Full method for equation of 2l . This involves an equation using the point (
24,0 and
M1
2
y
+
3
x
=
c
2
y
3
x
72
A1
+
=
(b)
M1(B1 on EPEN): Uses y = 0 in l1 in an attempt to find the x coordinate of B.
Checking any working for this mark but it may be implied by x = ±15
Correct attempt at area BPA using their 15
−
dM1
and 18. This requires
1
(
)
2 ×
24
+
''15''
×
''18''
or equivalent
1
(
)
2
2
(
)
2
2
correct work with their values e.g.
2 ×
24 12
−
+
18
×
12 15
+
+
18
Or e.g. “shoelace method”
−
15
24
12
−
15
1
1 0
=
+
24 18
×
+
0
−
0
−
0 15 18
+
×
2
0
0
18
0
2
Depends on the first mark and depends on their B being a point on the xaxis.
The working for the area takes precedence over any diagrams they have drawn so if the
working is correct but e.g. their diagram has points in the wrong positions, award the
marks.
There may be other methods e.g. Hero’s formula or use of trigonometry – if you are
unsure if such attempts deserve credit , use review.
A1
351 cso
Common errors seen in marking:
3
2
36 184
20
(
)
3
y
−
2
x
=
30
⇒
l
:
m
=
→
l
:
y
= −
x
−
24
→
P
,
,
B
−
, 0
1
2
2
3
13
13
3
2
3
276
−
54
(
)
(
)
3
y
−
2
x
=
30
⇒
l
:
m
= −
→
l
:
y
=
x
−
24
→
P
,
,
B
15, 0
3
2
13
13
1
2
With a correct triangle area method such attempts will often score (a) B0M1A1ftM1A0 (b)
M1M1A0
Triangle for reference:
(12, 18)
(−15, 0)
(24, 0)
39
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.