Question 1: [1 mark]
Define ‘Perpendicular’
Answer type: Multiple choice type 2
A: Lines that meet at 90 \degree.
B: Lines that pass through 0.
C: Lines that have the same gradient.
D: Lines that have the same y-intercept.
ANSWER: A: Lines that meet at 90 \degree.
Question 2: (Changed slightly to ‘parallel’ or ‘not parallel’ answers)
For each part (b) to (g), choose whether the lines are parallel to CD, perpendicular to CD, or neither.
Question 2(a): [2 marks]
The line CD is defined by the points C(-2,1) and D(10,7).
Choose the correct equation of the line CD.
Answer type: Multiple choice type 2
A: y = 2x - 13
B: y = \dfrac{1}{2} x - 2
C: y = \dfrac{3}{4} x - \dfrac{1}{2}
D: y = \dfrac{1}{2} x + 2
ANSWER: D: y = \dfrac{1}{2} x + 2
WORKING:
m = \dfrac{\text{change in} \, y}{\text{change in} \, x} = \dfrac{7 - 1}{10 - - 2} = \dfrac{6}{12} = \dfrac{1}{2}
7 = \dfrac{1}{2} \times 10 + c
7 = 5 + c
c = 2
y =\dfrac{1}{2} x + 2
Question 2(b): [1 mark]
y = - 2x
Answer type: Multiple choice type 2
A: Parallel
B: Perpendicular
C: Neither
ANSWER: B: Perpendicular
Question 2(c): [1 mark]
y = \dfrac{1}{2}x
Answer type: Multiple choice type 2
A: Parallel
B: Perpendicular
C: Neither
ANSWER: A: Parallel
Question 2(d): [1 mark]
12y = 6x + 7
Answer type: Multiple choice type 2
A: Parallel
B: Perpendicular
C: Neither
ANSWER: A: Parallel
Question 2(e): [1 mark]
2y = 2x + 2
Answer type: Multiple choice type 2
A: Parallel
B: Perpendicular
C: Neither
ANSWER: C: Neither
Question 2(f): [1 mark]
2(y - 3x) = 5 - 2x
Answer type: Multiple choice type 2
A: Parallel
B: Perpendicular
C: Neither
ANSWER: C: Neither
Question 2(g): [1 mark]
0 = \dfrac{2x + y}{2}
Answer type: Multiple choice type 2
A: Parallel
B: Perpendicular
C: Neither
ANSWER: B: Perpendicular
Question 3: [2 marks]
The line A is shown below.
Choose the equation below which is perpendicular to line A.
Answer type: Multiple choice type 2
A: y = \dfrac{2}{3} x + 4
B: y = - \dfrac{2}{3} x + 1
C: y = \dfrac{3}{2}x - 1
D: y = 3x
ANSWER: B: y = - \dfrac{2}{3} x + 1
WORKING:
The gradient of y = - \dfrac{2}{3} x + 1, - \dfrac{2}{3} is the negative reciprocal of the gradient of line A, which is \dfrac{3}{2}
Question 4:
A and B are two perpendicular lines with equations:
A: y = mx
B: y = px + 5
Question 4(a): [2 marks]
Choose the correct equation of line A.
Answer type: Multiple choice type 2
A: y = \dfrac{1}{2}x
B: y = - \dfrac{5}{4} x + 5
C: y = - 2x + 5
D: y = \dfrac{1}{2} x + 5
ANSWER: A: y = \dfrac{1}{2}x
WORKING:
\text{Gradient of} \, A = \dfrac{\text{change in} \, y}{\text{change in} \, x} = \dfrac{1}{2}
y = \dfrac{1}{2} x
Question 4(b): [2 marks]
Choose the correct equation of line B.
Answer type: Multiple choice type 2
A: y = \dfrac{1}{2}x
B: y = - \dfrac{5}{4} x + 5
C: y = - 2x + 5
D: y = \dfrac{1}{2} x + 5
ANSWER: C: y = - 2x + 5
WORKING:
\text{Gradient of} \, B = \dfrac{\text{change in} \, y}{\text{change in} \, x} = - \dfrac{2}{1} = -2
y = - 2x + 5
Question 5(a): [2 marks]
Choose the correct equation of the line that passes through (5, 4) and is perpendicular to y = -3x + 4.
Answer type: Multiple choice type 2
A: y = \dfrac{1}{3}x + \dfrac{7}{3}
B: y = \dfrac{1}{3}x + 3
C: y = - \dfrac{1}{3}x + \dfrac{7}{3}
D: y = - \dfrac{1}{3}x + 3
ANSWER: A: y = \dfrac{1}{3}x + \dfrac{7}{3}
WORKING:
Perpendicular so m \times - 3 = - 1 ; m = \dfrac{-1}{-3} ; m = \dfrac{1}{3}
Substituting values for x and y:
4 = \dfrac{1}{3} \times 5 + c
c = \dfrac{7}{3}
y = \dfrac{1}{3} + \dfrac{7}{3}
Question 5(b): [2 marks]
Choose the correct equation of the line that passes through (-1, -5) and is perpendicular to y = \dfrac{1}{3}x - 2.
Answer type: Multiple choice type 2
A: y = 3x - 2
B: y = 3x - 8
C: y = - 3x - 2
D: y = - 3x - 8
ANSWER: D: y = - 3x - 8
WORKING:
Perpendicular so m \times \dfrac{1}{3} = - 1 ; m = -1 \times 3 ; m = - 3
Substituting values for x and y:
- 5 = - 3 \times - 1 + c
c = - 8
y = - 3x - 8
Question 6(a): [3 marks]
The line A is given as 5y - 2x - 2 = 0.
The line B is perpendicular to A and passes through the point (1,-1)
Choose the correct x and y coordinates of the point of intersection of the two lines.
Answer type: Multiple choice type 1
A: x = \dfrac{11}{29}, y = \dfrac{16}{29}
B: x = \dfrac{16}{29}, y = \dfrac{11}{29}
C: x = \dfrac{10}{29}, y = \dfrac{16}{29}
D: x = \dfrac{11}{29}, y = \dfrac{10}{29}
ANSWER: A: x = \dfrac{11}{29}, y = \dfrac{16}{29}
WORKING:
Line A:
5y - 2x - 2 = 0
5y = 2x + 2
y = \dfrac{2}{5}x + \dfrac{2}{5}
Line B is perpendicular, so the gradient is:
m \times \dfrac{2}{5} = -1
m = - \dfrac{5}{2}
Equation of line B:
y = - \dfrac{5}{2}x + c
-1 = - \dfrac{5}{2} \times 1 + c
c = \dfrac{3}{2}
y = - \dfrac{5}{2}x + \dfrac{3}{2}
So we have:
y = \dfrac{2}{5} x + \dfrac{2}{5} ,
y = - \dfrac{5}{2}x + \dfrac{3}{2};
\dfrac{2}{5} x + \dfrac{2}{5} = - \dfrac{5}{2}x + \dfrac{3}{2}
\dfrac{29}{10} x = \dfrac{11}{10}
29 x = 11
x = \dfrac{11}{29}
Substituting this value back in to find y:
y = - \dfrac{5}{2}x + \dfrac{3}{2}
y = - \dfrac{5}{2} \times \dfrac{11}{29} + \dfrac{3}{2}
y = - \dfrac{55}{58} + \dfrac{3}{2}
y = \dfrac{16}{29}
Point of intersection is \bigg( \dfrac{11}{29}, \dfrac{16}{29} \bigg)
Question 6(b): [2 marks]
A third line, C, is perpendicular to B and has y-intercept of -3. Choose the correct equation of C.
Answer type: Multiple choice type 2
A: y = - \dfrac{5}{2}x + 3
B: y = - \dfrac{5}{2}x - 3
C: y = \dfrac{2}{5}x + 3
D: y = \dfrac{2}{5}x - 3
ANSWER: D: y = \dfrac{2}{5}x - 3
WORKING:
Has the same gradient as A, m = \dfrac{2}{5}.
Has a y – intercept of - 3, c = -3.
y = \dfrac{2}{5} x - 3