Question 8:
8 | (i) | AB2 = (1(1))2 + (5  1)2
BC2 = (3  (1))2 + (1 1)2
shown equal eg
AB2 = 22 + 42 [=20] and
BC2 = 42 + 22 [=20] with correct notation for
final comparison | M1
M1
A1
[3] | oe, or square root of this; condone poor
notation re roots; condone (1 + 1)2 instead of
(1(1))2
2
allow M1 for vector AB =  , condoning
4
poor notation, or triangle with hyp AB and
lengths 2 and 4 correctly marked
oe, or square root of this; condone poor
notation re roots; condone (3 + 1)2 instead of
(3(1))2 oe
 4 
allow M1 for vector BC =  , condoning
2
poor notation, or triangle with hyp BC and
lengths 4 and 2 correctly marked
or statement that AB and BC are each the
hypotenuse of a right-angled triangle with
sides 2 and 4 so are equal
SC2 for just AB2 = 22 + 42 and
BC2 = 42 + 22 (or roots of these) with no
clearer earlier working; condone poor
notation | eg A0 for AB = 20 etc
8 | (ii) | 51 6
[grad. of AC =] or oe
13 2
51 4
[grad. of BD =] or oe
111 12
showing or stating product of gradients = 1
or that one gradient is the negative reciprocal
of the other oe | M1
M1
B1
[3] | award at first step shown even if errors after
eg accept m × m = 1 or ‘one gradient is
1 2
negative reciprocal of the other’
B0 for ‘opposite’ used instead of ‘negative’
or ‘reciprocal’ | if one or both of grad AC = -3 and
grad BD = 1/3 seen without better
working for both gradients, award one
M1 only. For M1M1 it must be clear
that they are obtained independently
may be earned independently of
correct gradients, but for all 3 marks to
be earned the work must be fully
correct
8 | (iii) | midpoint E of AC = (2, 2) www
eqn BD is y 1x 4 oe
3 3
eqn AC is y = 3x + 8 oe
using both lines and obtaining intersection E
is (2, 2) (NB must be independently obtained
from midpt of AC)
midpoint F of BD = (5,3) | B1
M1
M1
A1
B1
[5] | condone missing brackets for both B1s
accept any correct form isw or correct ft their
gradients or their midpt F of BD
this mark will often be gained on the first
line of their working for BD
accept any correct form isw or correct ft their
gradients or their midpt E of AC
this mark will often be gained on the first
line of their working for AC
[see appendix for alternative methods instead
showing E is on BD for this M1]
this mark is often earned earlier
see the appendix for some common
alternative methods for this question; for all
methods, for A1 to be earned, all work for
the 5 marks must be correct | 0 for ((5+ 1)/2, (1+3)/2) = (2, 2)
may be earned using (2, 2) but then
must independently show that B or D
or (5, 3) is on this line to be eligible for
A1
if equation(s) of lines are seen in part
ii, allow the M1s if seen/used in this
part
[see appendix for alternative ways of
gaining these last two marks in
different methods]
for all methods show annotations M1
B1 etc then omission mark or A0 if
that mark has not been earned