Mathematics · Oct / Nov 2020 · Paper 11 · Question 6
YMA01
As printed in the paper
The question as text
6. The point A has coordinates (–4, 11) and the point B has coordinates (8, 2).
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[2]
2 11
−
11
−
2
E.g.
m
=
or
m
= −−
8
+
4
4
8
3
m = −
4
6(b)[4]
13
M is
2, 2
3
m
= −÷ −
1 "
"
N
4
13
4
(
)
y
−
"
"
=
"
"
x
−
"2"
2
3
8
x
−
6
y
+
23
=
0
6(c)[5]
(
)
(
) (
)
(
)
(
) (
)
AB =
−−
4
8
2
+
11
−
2
2
=
15
or
AB = −−
2
4
8
2
+
11 2
−
2
=
225
1
75
(
)
×
MC
×
AB
=
37.5
⇒
MC
=
=
5
or
MC =
2
25
2
15
4 ,
"
13
"
"
13
"
⇒
C
is "2" 3,
−
−
4
or
"2" 3,
+
+
4
m
=
MC
=
5
N
2
2
3
(
)
or (
)
−
1, 2.5
5, 10.5 or
x
= −
1,
x
=
5
or
y
=
2.5,
y
=
10.5
(
)
and (
)
−
1, 2.5
5, 10.5
Notes
(a)
M1 Correct gradient method. Method must be correct for both the numerator and denominator.
Could also be solved by setting up two simultaneous equations and solving correctly.
ie 11
= −
4
m
+
c
2
=
8
m
+
c
⇒
9
= −
12
m
⇒
m
=
... (M1)
Correct fraction or decimal (allow 3
−or 3
A1
− or −0.75). If they find the equation of the line then
4
4
3
they must identify
−
as the gradient.
4
(b)
−+
4
8
11
+
2
B1 Correct midpoint. Allow unsimplified e.g.
,
. Coordinates can be just stated,
2
2
seen in their working or they may appear on the diagram. Condone lack of brackets.
M1 Applies the perpendicular gradient rule to their gradient from part (a)
M1 Correct straight line method using their midpoint and a “changed” gradient. If using y = mx + c,
must reach as far as c = ….
A1 Allow any integer multiple
(c)
M1 Correct application of Pythagoras using the points A and B (or 1
2 AB using M)
M1 Uses their AB and 37.5 correctly to find MC or
MC . This may be implied by their working to
2
find eg AM or BM
dM1 Correct strategy to find both x coordinates, or both y coordinates or a coordinate pair for the
possible point C. To score this mark, look for sight of a linear or quadratic equation in one
variable that would find either an x or a y coordinate for one of the possible pairs of coordinates
for C. Condone slips in their rearrangements and they do not need to reach a value for either x or
y for this mark.
Usually look for:
Recognising that a distance of “5” for MC means by Pythagoras thatx =
"2" 3
± or
13
"
13
"
"
13
"
y =
"
" 4
±
. Alternatively,
C
is "2" 3,
−
−
4
or
"2" 3,
+
+
4
2
2
2
(If MC is not 5 look for application of Pythagoras.)
Forming an equation of a circle with radius “5” from a point C and solvingsimultaneously with their l eg
(
x
−
"2")
2
+
(
y
−
"
")
2
=
"25"
and "8
x
−
6
y
+
23
=
0"
⇒
(
x
−
2)
2
+
(
x
+
−
13
4
23
13
2
3
6
2
(they may form an equation in y).
Shoelace method:−
4
11
x
4
x
+
23
(
)
3
6
=
1
−
4(
4
x
+
23
)
+
2
x
+
88)
−
(11
x
+
8(
4
x
+
23
)
−
8)
=
37.5
8
2
2
3
6
3
6
−
4
11
It is dependent on both of the previous method marks. There may be other credit worthy
methods so if unsure then send to review.
A1
( 1, 2.5)
−
or (5, 10.5) (does not have to be written as coordinates)
or
x
= −
1,
x
=
5
or
y
=
2.5,
y
=
10.5
A1
( 1, 2.5)
−
and (5, 10.5). Condone lack of brackets.
They may state the values for x and y but the coordinates must be paired correctly.
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