5.15H Equation of a Line: Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Work out the gradient of the line joining the points (2, 3) and (5, 11).

(Total for Question 1 is 1 mark)

Find the equation of the line with gradient -2 that passes through the point (4, -1). Give your answer in the form y = mx + c.

(Total for Question 2 is 3 marks)

Give the equation of the line in standard form ax + by + c = 0 that passes through the points (0, 5) and (3, -1).

(Total for Question 3 is 3 marks)

The line L has equation 3x - 2y + 6 = 0.
(a) Work out the gradient of L.
(b) Find the equation of the line through (2, 1) parallel to L. Give your answer in the form y = mx + c.
(c) Find the equation of the line through (2, 1) perpendicular to L. Give your answer in the form y = mx + c.

(a)Work out the gradient of L.(1)

(b)Find the equation of the line through (2, 1) parallel to L. Give your answer in the form y = mx + c.(2)

(c)Find the equation of the line through (2, 1) perpendicular to L. Give your answer in the form y = mx + c.(1)

(Total for Question 4 is 4 marks)

Work out the intersection point of the lines y = 2x + 1 and y = -x + 7. Give full working and the final coordinates.

(Total for Question 5 is 5 marks)

Show that the points A(1, 2), B(3, -2) and C(5, -6) are collinear. State the equation of the line containing them.

(Total for Question 6 is 4 marks)

Find the equation of the perpendicular bisector of the line segment joining (1, 1) and (5, 3). Give your answer in the form y = mx + c.

(Total for Question 7 is 5 marks)

Find the value of k for which the line y = kx + 3 is perpendicular to the line 2x + y - 5 = 0. Show your working.

(Total for Question 8 is 3 marks)

Mark scheme · 5.15H Equation of a Line: Higher Tier Practice

Question 1

B1 8/3 cao
Answer: 8/3

Question 2

M1 substitute point into y = -2x + c to form an equation for c
A1 c = 7 cao
A1 y = -2x + 7 cao
Answer: y = -2x + 7

Question 3

M1 find gradient m = (-1 - 5)/(3 - 0) = -6/3 = -2
M1 use y = mx + c or point-substitution to find c = 5
A1 2x + y - 5 = 0 cao
Answer: 2x + y - 5 = 0

Question 4

(a) B1 3/2 cao
(a) Answer: 3/2
(b) M1 use gradient 3/2 and substitute (2,1) into y = 3/2 x + c to find c
(b) A1 y = 3/2 x - 2 cao
(b) Answer: y = 3/2 x - 2
(c) A1 y = -2/3 x + 7/3 cao
(c) Answer: y = -2/3 x + 7/3

Question 5

M1 equate 2x + 1 = -x + 7 to form an equation in x
M1 collect terms and solve for x giving x = 2
A1 x = 2 cao
M1 substitute x = 2 into one of the equations to find y
A1 y = 5 cao
Answer: (2, 5)

Question 6

M1 find gradient AB = (-2 - 2)/(3 - 1) = -4/2 = -2
M1 find gradient BC = (-6 - (-2))/(5 - 3) = -4/2 = -2
A1 state gradients equal, so collinear cao
B1 give equation y = -2x + 4 cao
Answer: They are collinear; equation y = -2x + 4

Question 7

M1 find midpoint ( (1+5)/2, (1+3)/2 ) = (3, 2 )
M1 find gradient of segment = (3 - 1)/(5 - 1) = 2/4 = 1/2
M1 perpendicular gradient = -2
M1 form equation y - 2 = -2(x - 3)
A1 y = -2x + 8 cao
Answer: y = -2x + 8

Question 8

M1 rearrange 2x + y - 5 = 0 to y = -2x + 5
M1 use product of gradients k * -2 = -1 for perpendicular lines
A1 k = 1/2 cao
Answer: 1/2

Equation of a Line: Higher Tier Practice - Worksheets, Questions and Revision