Mathematics · Oct / Nov 2024 · Paper 11 · Question 1
YMA01
As printed in the paper
The question as text
1. The line l1 passes through the point A(–5, 20) and the point B(3, –4).
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
1(a)[3]
( 3)
−−
5
3
Attempts equation of line
y
−
20
= −
" 3"(
x
+
5)
or
y
+
4
= −
" 3"(
x
−
3)
y
3
x
+
5
= −
3 or midpoint (
)
Gradient 1
−
1, 8
1(b)[3]
1
y
−
8
=
(
x
+
1)
3
x
−
3
y
+
25
=
0
(a)
M1: Attempts gradient = y
condoning slips. Look for an attempt at a difference in y
x
coordinates over a difference in x coordinates. There is no requirement to simplify.
(
)
They may use
20
−−=
4
m
−−
5
3
to find m.
dM1: Full method for equation of line using gradient and either point – as shown or may use
" 3"
y
= −
x
+
c
proceeding as far as finding c.
24
3
5
A1:
y
= −
x
+
The coefficients must be simplified, award A0 for
y
=
x
+
5
−
8
Alt (a)
M1: Sets up two simultaneous equations using both points E.g 20
= −
5m
+
c
and
−=
4
3m
+
c
(allow one slip).
dM1: Solves the pair of simultaneous equations to find values for m and c (accept any values
following two suitable equations having been set up).
A1:
y
= −
3
x
+
5
(b)
B1ft: Correct mid point or gradient, allowing follow through on the negative reciprocal of their
gradient in (a)
M1: Attempts the perpendicular bisector (need not be a simplified form). Must be using ±
the reciprocal of their gradient from (a) with a valid attempt at the midpoint – at least
one correct coordinate (or calculation for) must be seen for the midpoint.
A1:
x
−
3
y
+
25
=
0
but allow any integer multiple and term in a different order (but must be
3
25
0
on one side). Ignore if they give a label 2
l
=
x
−
y
+
=
3
3
4
3
y
2
x
10
6
x
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