Question 5:
5 | (i) |  1 5 
midpt of AB =  ,  oe www
 2 2 
4 1
grad AB = oe
3  (2)
using gradient of AB to obtain grad perp
bisector
5
y  2.5 = (x  0.5) oe
3 | B2
M1
M1
M1 | allow unsimplified
B1 for one coordinate correct
must be obtained independently of given line;
accept 3 and 5 correctly shown eg in a sketch,
followed by 3/5
M1 for rise/run = 3/5 etc
M0 for just 3/5 with no evidence
for use of m m = 1 soi or ft their gradient
1 2
AB
5
M0 for just without AB grad found
3
5
eg M1 for y  x  c and subst of midpt;
3
ft their gradient of perp bisector and midpt;
M0 for just rearranging given equation | if working shown, should come from
 3  2 4 1
 ,  oe
 2 2 
5
NB B0 for x coord. = , (obtained
2
from subtraction instead of addition)
for those who find eqn of AB first, M0
y  4 x  3
for just  oe, but M1 for
1 4 2  3
1 4
y  4  x  3 oe
2  3
ignore their going on to find the eqn of
AB after finding grad AB
this second M1 available for starting
5
with given line = and obtaining
3
grad. of AB from it
no ft for gradient of AB used
completion to given answer 3y + 5x = 10,
showing at least one interim step | M1
[6] | condone a slight slip if they recover quickly
and general steps are correct ( eg sometimes a
5
slip in working with the c in y  x  c
3
-condone 3y = 5x + c followed by
substitution and consistent working)
M0 if clearly ‘fudging' | NB answer given; mark process not
answer; annotate if full marks not
earned eg with a tick for each mark
earned
scores such as B2M0M0M1M1 are
possible
after B2, allow full marks for complete
method of showing given line has
gradient perp to AB (grad AB must be
found independently at some stage) and
passes through midpt of AB
5 | (ii) | 3y + 5(4y  21) = 10
(1, 5) or y = 5, x = 1 isw | M1
A2
[3] | or other valid strategy for eliminating one
5 10 x 21
variable attempted eg x    ;
3 3 4 4
condone one error
A1 for each value;
if AO allow SC1 for both values correct but
 23 115 
unsimplified fractions, eg  , 
 23 23  | or eg 20y = 5x + 105 and subtraction of
two eqns attempted
no ft from wrong perp bisector eqn,
since given
allow M1 for candidates who reach
y = 115/23 and then make a worse
attempt, thinking they have gone wrong
NB M0A0 in this part for finding E
using info from (iii) that implies E is
midpt of CD
5 | (iii) | (x  a)2 + (y  b)2 = r2 seen or used
12 + 42 oe (may be unsimplified), from clear
use of A or B
(x + 1)2 + (y  5)2 = 17
showing midpt of CD = (1, 5)
showing CE or DE = 17 oe or showing one
of C and D on circle | M1
M1
A1
M1
M1 | or for (x + 1)2 + (y  5)2 =k, or ft their E,
where k > 0
for calculating AE or BE or their squares, or
for subst coords of A or B into circle eqn to
find r or r2, ft their E;
for eqn of circle centre E, through A and B;
allow A1 for r2 = 17 found after
(x + 1)2 + (y  5)2 = r2 stated and second M1
clearly earned
if (x + 1)2 + (y  5)2 = 17 appears without
clear evidence of using A or B, allow the first
M1 then M0 SC1
alt M1 for showing CD2 = 68 oe
allow to be earned earlier as an invalid
attempt to find r | this M not earned for use of CE or DE
or ½ CD
NB some cands finding AB2 = 34 then
obtaining 17 erroneously so M0
SC also earned if circle comes from C
or D and E, but may recover and earn
the second M1 later by using A or B
[5] | showing that both C and D are on circle and
commenting that E is on CD is enough for
last M1M1;
similarly showing CD2 = 68 and both C and
D are on circle oe earns last M1M1 | other methods exist, eg: may find eqn
of circle with centre E and through C or
D and then show that A and B and
other of C/D are on this circle – the
marks are then earned in a different
order; award M1 for first fact shown
and then final M1 for completing the
argument;
if part-marks earned, annotate with a
tick for each mark earned beside where
earned