Question
P (4, 1) and Q (0, –5) are points on the coordinate plane.
a.Determine the
(i) coordinates of M, the midpoint of P and Q.
(ii) gradient of the line drawn through P and Q.
(iii) gradient of the line drawn through M, perpendicular to PQ.[4]
Find k.[2]
▶️Answer/Explanation
Markscheme
(i) (2, – 2) parentheses not required. (A1)
(ii) gradient of PQ \( = \left( {\frac{{ – 5 – 1}}{{0 – 4}}} \right) = \frac{6}{4} = \frac{3}{2}(1.5)\) (M1)(A1)
(M1) for gradient formula with correct substitution
Award (A1) for \(y = \frac{3}{2}x – 5\) with no other working
(iii) gradient of perpendicular is \( – \frac{2}{3}\) (A1)(ft) (C4)[4 marks]
\(\left( {\frac{{k + 2}}{{0 – 2}}} \right) = – \frac{2}{3}\), \(k = – \frac{2}{3}\) or \(y = – \frac{2}{3}x + c\), \(c = – \frac{2}{3}\therefore k = – \frac{2}{3}\) (M1)(A1)(ft)
Allow (\(0, – \frac{2}{3}\))
(M1) is for equating gradients or substituting gradient into \(y = mx + c\) (C2)[2 marks]
Question
A store sells bread and milk. On Tuesday, 8 loaves of bread and 5 litres of milk were sold for $21.40. On Thursday, 6 loaves of bread and 9 litres of milk were sold for $23.40.
If \(b =\) the price of a loaf of bread and \(m =\) the price of one litre of milk, Tuesday’s sales can be written as \(8b + 5m = 21.40\).
a.Using simplest terms, write an equation in b and m for Thursday’s sales.[2]
b.Find b and m.[2]
c.Draw a sketch, in the space provided, to show how the prices can be found graphically.
[2]
▶️Answer/Explanation
Markscheme
Thursday’s sales, \(6b + 9m = 23.40\) (A1)
\(2b + 3m = 7.80\) (A1) (C2)[2 marks]
\(m = 1.40\) (accept 1.4) (A1)(ft)
\(b = 1.80\) (accept 1.8) (A1)(ft)
Award (A1)(d) for a reasonable attempt to solve by hand and answer incorrect. (C2)[2 marks]
(A1)(A1)(ft)
(A1) each for two reasonable straight lines. The intersection point must be approximately correct to earn both marks, otherwise penalise at least one line.
Note: The follow through mark is for candidate’s line from (a). (C2)[2 marks]
Question
The mid-point, M, of the line joining A(s , 8) to B(−2, t) has coordinates M(2, 3).
a.Calculate the values of s and t.[2]
b.Find the equation of the straight line perpendicular to AB, passing through the point M.[4]
▶️Answer/Explanation
Markscheme
\(s = 6\) (A1)
\(t = – 2\) (A1) (C2)[2 marks]
\({\text{gradient of AB}} = \frac{{ – 2 – 8}}{{ – 2 – 6}} = \frac{{ – 10}}{{ – 8}} = \frac{5}{4}\) (A1)(ft)
(A1) for gradient of AM or BM \( = \frac{5}{4}\)
\({\text{Perpendicular gradient}} = – \frac{4}{5}\) (A1)(ft)
Equation of perpendicular bisector is
\(y = – \frac{4}{5}x + c\)
\(3 = – \frac{4}{5}(2) + c\) (M1)
\(c = 4.6\)
\(y = -0.8x + 4.6\)
or \(5y = -4x + 23\) (A1)(ft) (C4)[4 marks]
Question
a.Write down the gradient of the line \(y = 3x + 4\).[1]
b.Find the gradient of the line which is perpendicular to the line \(y = 3x + 4\).[1]
c.Find the equation of the line which is perpendicular to \(y = 3x + 4\) and which passes through the point \((6{\text{, }}7)\).[2]
d.Find the coordinates of the point of intersection of these two lines.[2]
▶️Answer/Explanation
Markscheme
\(3\) (A1) (C1)[1 mark]
\( – 1/3\) (ft) from (a) (A1)(ft) (C1)[1 mark]
Substituting \((6{\text{, }}7)\) in \(y ={\text{their }}mx + c\) or equivalent to find \(c\). (M1)
\(y = \frac{{ – 1}}{3}x + 9\) or equivalent (A1)(ft) (C2)[2 marks]
\((1.5{\text{, }}8.5)\) (A1)(A1)(ft) (C2)
Note: Award (A1) for \(1.5\), (A1) for \(8.5\). (ft) from (c), brackets not required.[2 marks]
Question
A straight line, \({L_1}\) , has equation \(x + 4y + 34 = 0\) .
a.Find the gradient of \({L_1}\) .[2]
b.The equation of line \({L_2}\) is \(y = mx\) . \({L_2}\) is perpendicular to \({L_1}\) .
Find the value of \(m\).[2]
c.The equation of line \({L_2}\) is \(y = mx\) . \({L_2}\) is perpendicular to \({L_1}\) .
Find the coordinates of the point of intersection of the lines \({L_1}\) and \({L_2}\) .[2]
▶️Answer/Explanation
Markscheme
\(4y = – x – 34\) or similar rearrangement (M1)
\({\text{Gradient}} = – \frac{1}{4}\) (A1) (C2)
\(m = 4\) (A1)(ft)(A1)(ft) (C2)
Note: (A1) Change of sign
(A1) Use of reciprocal[2 marks]
Reasonable attempt to solve equations simultaneously (M1)
\(( – 2{\text{, }} – 8)\) (A1)(ft) (C2)
Note: Accept \(x = – 2\) \(y = – 8\). Award (A0) if brackets not included.[2 marks]