Mathematics · May / June 2021 · Paper 11 · Question 7
YMA01
As printed in the paper
The question as text
7. y l2 B C 1.8 rad D O x E l1 Figure 3 The line l1 has equation 4y + 3x = 48 The line l1 cuts the y-axis at the point C, as shown in Figure 3.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[1]
y
coordinate
=
12
7(b)[3]
3
Gradient of l1 =
−
4
⇒Gradient of l2 = 4
4
3 ⇒(
)
y
−
6
=
" "(
x
−
8)
3
14
y coordinate =
−
*
3
7(c)[3]
14
50
Radius
=
"12"
+
=
3
3
50
Length of arc
=
"
" 1.8
×
=
30
3
7(d)[3]
1
50
(
)
Area of sector =
×
("
")
2
×
1.8
=
250
2
3
1
50
200
"250"
"
" 8
"250" "
"
+
×
× =
+
2
3
3
950
(units2)
=
3
Mark all parts together. May work in degrees.
(a)
B1
12 (Check by the question and also on the diagram). If there is a contradiction then their
answer in the main solution takes precedence.
(b)
3
but can be implied by further work. Eg sight of a gradient of 4
B1
States gradient of l1 is
−
4
in their equation for l2 can also score this mark.
The value must be identified or used so it cannot just be awarded from a rearranged equation
3
for l1. Circling the coefficient is acceptable but stating
−
4 x
with no further work is B0.
3
4
M1
Attempts to find the gradient of the perpendicular line
"
−
"
→
and attempts to find the
4
3
4
equation of l2. Look for (
)
y
−
6
=
" "(
x
−
8)
with both of the brackets correct. If they
3
attempt using y
+ then they must proceed as far as
=
mx
c
c =
...
14
A1*
−
cso must be clearly stated as the y coordinate with no errors seen after achieving a
3
correct equation for l2.
(c)
14
B1ft Finds the radius of the circle following through on their answer to (a).
"12"
+
is
3
acceptable for this mark or it may be implied by their length of the arc. May be seen on the
diagram or in other parts.
14
M1
Attempts to find the length of the arc with
θ =
1.8 and their
r =
"12"
+
3
A1
30 cao
(d)
14
M1
Attempts to find the area of the sector with
their
r =
"12"
+
θ =
1.8 and
3
M1
Adds the area of their sector with a correct method to find the area of the triangle.
There are various ways to find the area of the triangle. They may find the lengths CD and
DE using Pythagoras and proceed to find the area of the triangle:
32
2
40
1
40
200
Eg
and
CD =
6
2
+
8
2
=
10
DE
=
8
2
+
=
⇒
Area
=
×
10
×
=
3
3
2
3
3
Alternatively, via the shoelace method:
0 12
1
8 6
1
14
200
Eg
(
)
=
×
8
× −
−
12 8
×
=
0
2
−
14
2
3
3
3
0 12
A1 950
316 or 316.6& but not 316.7)
3 cao (accept
2
3
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