Mathematics · May / June 2025 · Paper 11 · Question 5
YMA01
As printed in the paper
The question as text
5. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. The line l1 has equation x – 2y + 25 = 0 The line l2 passes through the origin and is perpendicular to l1
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[2]
y
=
x
+
y
= −
x
+
2
2
y
= −
2
x
Substitutes
y
= −
2
x
into
x
−
2
y
+
25
=
0
x
+
4
x
+
25
=
0
5(b)[3]
x
5,
y
10
= −
=
5(c)[2]
2
2
Shortest distance =
5
+
10
=
5 5
(a)
M1: Attempts the normal gradient. E.g. writes
x
−
2
y
+
25
=
0
in the form y
=
px
+
q
AND changes
the gradient (need not be correctly changed). Alternatively, may switch the x and y coordinates
and change a sign – e.g. perpendicular line is 2
x
+
y
+
k
=
0
. Another approach is to find the
gradient of l1 by identifying two points on the line and attempting the gradient (must be change in
y over change in x) between these two points.
May be implied by the correct answer.
A1:
2
o.e. Score for the correct answer even if an incorrect “q” was seen.
y
= −
x
(b)
M1: Attempts to solve their
2
(must be of form kx) with
2
25
0
(or their rearrangement
y
= −
x
x
−
y
+
=
of this).
Score for setting up an equation in a single variable.
A1:
x = − OR
5
y =
10
from correct work.
A1: Both
from correct work
x
= −
5,
y
=
10
(c)
M1: Attempts to use Pythagoras theorem to find the distance or distance squared between (0, 0) and
their ( 5,10)
−
or other full method to find the shortest distance.
A1cso: 5 5 Must be simplified surd form. Must have come from correct coordinates in (b).
sin
CAO
sin 0.6
=
CAO
=
0.6944...
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.