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8.9 Perpendicular Lines and the equation of a tangent

EDEXCEL 1MA1 · Calculator allowed · about 115 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

For each gradient given below, write down the gradient of a line that is perpendicular to it.

(a)Gradient = 2(1)

(b)Gradient = -4(1)

(c)Gradient = 2/5(1)

(Total for Question 1 is 3 marks)

Which of these lines is perpendicular to y = -2x + 5?
A) y = 2x - 3
B) y = (1/2)x + 1
C) y = -(1/2)x + 4
D) y = -2x + 7

A) y = 2x - 3
B) y = (1/2)x + 1
C) y = -(1/2)x + 4
D) y = -2x + 7

(Total for Question 2 is 1 mark)

A(2, 5) and B(6, -3) are two points on a coordinate grid.
Show that the line through A and B is perpendicular to the line with equation y = (1/2)x + 4

(Total for Question 3 is 3 marks)

Line L has equation 3x + y = 7
Find the equation of the line that is perpendicular to L and passes through the point (-3, 2). Give your answer in the form x + by = c, where b and c are integers.

(Total for Question 4 is 4 marks)

A circle has centre O, the origin. For each point P given below, P lies on the circle, and OP is a radius.
Without finding the full equation of the tangent, state the gradient of the tangent to the circle at P.

(a)P = (4, 2)(2)

(b)P = (-6, 3)(2)

(c)P = (-5, -2)(2)

(Total for Question 5 is 6 marks)

A circle with centre O, the origin, has equation x² + y² = 25
P is the point (3, 4).

(a)Show that P lies on the circle.(2)

(b)Find the equation of the tangent to the circle at P. Give your answer in the form ax + by = c, where a, b and c are integers.(3)

(Total for Question 6 is 5 marks)

A circle with centre O, the origin, has equation x² + y² = 50
The point P(-5, 5) lies on the circle.
Find the equation of the tangent to the circle at P. Give your answer in the form y = mx + c

(Total for Question 7 is 4 marks)

A circle with centre O, the origin, has equation x² + y² = 34
The point (5, 3) lies on the circle.
Find the equation of the tangent to the circle at this point. Give your answer in the form ax + by = c, where a, b and c are integers.

(Total for Question 8 is 4 marks)

A circle with centre O, the origin, has equation x² + y² = 169
The point (12, 5) lies on the circle.
Find the equation of the tangent to the circle at this point. Give your answer in the form ax + by = c, where a, b and c are integers.

(Total for Question 9 is 4 marks)

A circle has centre C(2, 3) and radius 5, so it has equation (x - 2)² + (y - 3)² = 25
P is the point (5, 7).

(a)Show that P lies on the circle.(2)

(b)Find the equation of the tangent to the circle at P. Give your answer in the form ax + by = c, where a, b and c are integers.(4)

(Total for Question 10 is 6 marks)

A circle has centre C(-1, 2) and radius √20, so it has equation (x + 1)² + (y - 2)² = 20
P is the point (3, 4).

(a)Show that P lies on the circle.(2)

(b)Find the equation of the tangent to the circle at P. Give your answer in the form y = mx + c(4)

(Total for Question 11 is 6 marks)

A circle with centre O, the origin, has equation x² + y² = 169
P is a point on the circle in the fourth quadrant (so the x-coordinate of P is positive and the y-coordinate of P is negative). The x-coordinate of P is 5.

(a)Find the y-coordinate of P.(2)

(b)Find the equation of the tangent to the circle at P. Give your answer in the form ax + by = c, where a, b and c are integers.(4)

(Total for Question 12 is 6 marks)

A circle with centre O, the origin, has equation x² + y² = 100
The tangent to the circle at the point P(6, 8) crosses the x-axis at point A and crosses the y-axis at point B.
Diagram: coordinate grid showing the circle x² + y² = 100, the tangent line at P(6, 8), and the points A and B where the tangent meets the axes, with triangle OAB formed.

Diagram NOT accurately drawn

(a)Find the equation of the tangent to the circle at P.(3)

(b)Find the coordinates of A.(1)

(c)Find the coordinates of B.(1)

(d)Calculate the area of triangle OAB, where O is the origin. Give your answer as an exact fraction or correct to 3 significant figures.(3)

(Total for Question 13 is 8 marks)

A circle with centre O, the origin, has equation x² + y² = 20
P is the point (4, 2) and Q is the point (-4, -2). Both P and Q lie on the circle.
Prove that the tangent to the circle at P is parallel to the tangent to the circle at Q.

(Total for Question 14 is 4 marks)

Priya is designing a circular pond in her garden. On a coordinate grid measured in metres, the pond has centre O, the origin, and its edge passes through the point P(9, 12). A straight path is to be built along the tangent to the pond's edge at P.

(a)Find the equation of the path. Give your answer in the form ax + by = c, where a, b and c are integers.(4)

(b)The main garden path runs along the x-axis. Find the coordinates of the point where Priya's new path crosses the main garden path.(2)

(Total for Question 15 is 6 marks)

A circle with centre O, the origin, has equation x² + y² = 25
P is the point (-3, 4).

(a)Find the equation of the tangent to the circle at P.(4)

(b)This tangent intersects the line with equation y = 2x + 1 at a single point. Find the coordinates of this point.(3)

(Total for Question 16 is 7 marks)

A circle with centre O, the origin, has equation x² + y² = r²
P is the point (a, b), where P lies on the circle and b is not 0.

(a)Prove that the gradient of the tangent to the circle at P is -a/b.(3)

(b)Hence show that the equation of the tangent to the circle at P(a, b) can be written as ax + by = r²(3)

(Total for Question 17 is 6 marks)

A circle with centre O, the origin, has equation x² + y² = 169
The line with equation 5x + 12y = 169 is a tangent to this circle.
Show that the point of contact of the tangent with the circle is (5, 12).

(Total for Question 18 is 4 marks)

A circle passes through the points A(1, 7), B(9, 7) and C(9, -1).
Diagram: coordinate grid showing points A(1, 7), B(9, 7) and C(9, -1) plotted, with a circle drawn through all three points.

Diagram NOT accurately drawn

(a)Find the equation of the perpendicular bisector of AB.(2)

(b)Find the equation of the perpendicular bisector of BC.(2)

(c)Hence find the coordinates of the centre of the circle.(2)

(d)Find the radius of the circle, and hence write down the equation of the circle.(3)

(Total for Question 19 is 9 marks)

Mark scheme · 8.9 Perpendicular Lines and the equation of a tangent

Question 1

(a) B1 -1/2 (oe -0.5) (cao)
(a) Answer: -1/2
(b) B1 1/4 (oe 0.25) (cao)
(b) Answer: 1/4
(c) B1 -5/2 (oe -2.5) (cao)
(c) Answer: -5/2

Question 2

B1 B selected (cao)
Answer: B (y = (1/2)x + 1)

Question 3

M1 correct gradient of AB, e.g. (-3 - 5)/(6 - 2) = -2 (oe)
M1 multiplies gradient of AB by 1/2 (the gradient of the given line)
C1 correct conclusion with full reasoning: (-2) x (1/2) = -1, so the lines are perpendicular (cso)
Answer: Perpendicular, since gradient AB x gradient of given line = -1

Question 4

M1 rearranges L into the form y = mx + c and identifies its gradient as -3
M1 finds the perpendicular gradient as 1/3
M1 correct method using the point (-3, 2), e.g. y - 2 = (1/3)(x + 3)
A1 x - 3y + 9 = 0 (oe, e.g. x - 3y = -9) (cao)
Answer: x - 3y + 9 = 0

Question 5

(a) M1 gradient of OP = 2/4 = 1/2 (oe)
(a) A1 -2 (cao)
(a) Answer: -2
(b) M1 gradient of OP = 3/(-6) = -1/2 (oe)
(b) A1 2 (cao)
(b) Answer: 2
(c) M1 gradient of OP = -2/-5 = 2/5 (oe)
(c) A1 -5/2 (oe -2.5) (cao)
(c) Answer: -5/2

Question 6

(a) M1 substitutes x = 3 and y = 4 into x² + y²
(a) A1 3² + 4² = 9 + 16 = 25, which matches the circle equation (cso)
(a) Answer: 3² + 4² = 25 (shown)
(b) M1 gradient of OP = 4/3, so tangent gradient = -3/4 (using tangent perpendicular to radius)
(b) M1 correct method using point P, e.g. y - 4 = -(3/4)(x - 3)
(b) A1 3x + 4y = 25 (oe, integer coefficients) (cao)
(b) Answer: 3x + 4y = 25

Question 7

M1 gradient of OP = 5/(-5) = -1 (oe)
M1 tangent gradient = 1 (negative reciprocal of -1)
M1 correct method using point P, e.g. y - 5 = 1(x + 5)
A1 y = x + 10 (cao)
Answer: y = x + 10

Question 8

M1 gradient of radius = 3/5
M1 tangent gradient = -5/3 (negative reciprocal of 3/5)
M1 correct method using the point (5, 3), e.g. y - 3 = -(5/3)(x - 5)
A1 5x + 3y = 34 (oe, integer coefficients) (cao)
Answer: 5x + 3y = 34

Question 9

M1 gradient of radius = 5/12
M1 tangent gradient = -12/5 (negative reciprocal of 5/12)
M1 correct method using the point (12, 5), e.g. y - 5 = -(12/5)(x - 12)
A1 12x + 5y = 169 (oe, integer coefficients) (cao)
Answer: 12x + 5y = 169

Question 10

(a) M1 substitutes x = 5 and y = 7 into (x - 2)² + (y - 3)²
(a) A1 (5 - 2)² + (7 - 3)² = 9 + 16 = 25, which matches the circle equation (cso)
(a) Answer: 3² + 4² = 25 (shown)
(b) M1 gradient of CP = (7 - 3)/(5 - 2) = 4/3 (oe)
(b) M1 tangent gradient = -3/4 (negative reciprocal of 4/3)
(b) M1 correct method using point P, e.g. y - 7 = -(3/4)(x - 5)
(b) A1 3x + 4y = 43 (oe, integer coefficients) (cao)
(b) Answer: 3x + 4y = 43

Question 11

(a) M1 substitutes x = 3 and y = 4 into (x + 1)² + (y - 2)²
(a) A1 (3 + 1)² + (4 - 2)² = 16 + 4 = 20, which matches the circle equation (cso)
(a) Answer: 4² + 2² = 20 (shown)
(b) M1 gradient of CP = (4 - 2)/(3 - (-1)) = 1/2 (oe)
(b) M1 tangent gradient = -2 (negative reciprocal of 1/2)
(b) M1 correct method using point P, e.g. y - 4 = -2(x - 3)
(b) A1 y = -2x + 10 (cao)
(b) Answer: y = -2x + 10

Question 12

(a) M1 substitutes x = 5 into x² + y² = 169 to get 25 + y² = 169, so y² = 144
(a) A1 y = -12, selecting the negative root as P is in the fourth quadrant (cao)
(a) Answer: y = -12
(b) M1 gradient of OP = -12/5, ft their part (a)
(b) M1 tangent gradient = 5/12 (negative reciprocal of -12/5)
(b) M1 correct method using point P, e.g. y + 12 = (5/12)(x - 5)
(b) A1 5x - 12y = 169 (oe, integer coefficients) (cao)
(b) Answer: 5x - 12y = 169

Question 13

(a) M1 gradient of OP = 8/6 = 4/3, so tangent gradient = -3/4
(a) M1 correct method using point P, e.g. y - 8 = -(3/4)(x - 6)
(a) A1 y = -(3/4)x + 12.5 (oe, e.g. 3x + 4y = 50) (cao)
(a) Answer: y = -(3/4)x + 12.5
(b) A1 (50/3, 0) (oe awrt (16.7, 0)), ft their part (a)
(b) Answer: A = (50/3, 0)
(c) A1 (0, 12.5) (oe (0, 25/2)), ft their part (a)
(c) Answer: B = (0, 12.5)
(d) M1 recognises that OA and OB are perpendicular (along the axes), so area = (1/2) x OA x OB
(d) M1 substitutes their values, e.g. (1/2) x (50/3) x 12.5
(d) A1 625/6 (oe awrt 104) square units, ft their parts (b) and (c)
(d) Answer: Area = 625/6 square units (awrt 104 square units)

Question 14

M1 gradient of OP = 2/4 = 1/2, so tangent gradient at P = -2
M1 gradient of OQ = (-2)/(-4) = 1/2, so tangent gradient at Q = -2
A1 both tangent gradients equal -2
C1 correct conclusion: since the tangents have equal gradients and touch the circle at different points, they are parallel (not the same line), as required (cso)
Answer: Proved: both tangents have gradient -2, so they are parallel

Question 15

(a) M1 recognises OP = √9² + 12² = 15, so the equation of the pond's edge is x² + y² = 225
(a) M1 gradient of OP = 12/9 = 4/3, so tangent gradient = -3/4
(a) M1 correct method using point P, e.g. y - 12 = -(3/4)(x - 9)
(a) A1 3x + 4y = 75 (oe, integer coefficients) (cao)
(a) Answer: 3x + 4y = 75
(b) M1 sets y = 0 in their equation from part (a), ft
(b) A1 (25, 0) (cao, ft)
(b) Answer: (25, 0)

Question 16

(a) M1 gradient of OP = 4/(-3) = -4/3, so tangent gradient = 3/4
(a) M1 correct method using point P, e.g. y - 4 = (3/4)(x + 3)
(a) M1 correct rearrangement towards integer form
(a) A1 -3x + 4y = 25 (oe, e.g. y = (3/4)x + 25/4) (cao)
(a) Answer: -3x + 4y = 25 (oe y = (3/4)x + 25/4)
(b) P1 sets their tangent equation equal to 2x + 1, ft their part (a)
(b) M1 correct algebraic method to solve for x, e.g. (3/4)x + 25/4 = 2x + 1 leading to 3x + 25 = 8x + 4
(b) A1 (21/5, 47/5) (oe (4.2, 9.4)), both coordinates correct (cao)
(b) Answer: (21/5, 47/5) = (4.2, 9.4)

Question 17

(a) M1 states the gradient of the radius OP is b/a (a not 0)
(a) M1 uses the perpendicular gradient rule: (gradient of OP) x (gradient of tangent) = -1
(a) C1 correct conclusion: gradient of tangent = -1 / (b/a) = -a/b, as required (cso)
(a) Answer: Proved: tangent gradient = -a/b
(b) M1 uses the point-gradient form with their result from part (a), e.g. y - b = -(a/b)(x - a)
(b) M1 multiplies both sides by b and expands: by - b² = -ax + a²
(b) C1 correctly rearranges to ax + by = a² + b², then uses a² + b² = r² (since P lies on the circle) to conclude ax + by = r² (cso)
(b) Answer: Proved: ax + by = r²

Question 18

B1 checks 5² + 12² = 169, so (5, 12) lies on the circle
B1 checks 5(5) + 12(12) = 169, so (5, 12) lies on the tangent line
M1 checks perpendicularity: gradient of the radius to (5, 12) is 12/5, and the gradient of the line 5x + 12y = 169 is -5/12, and (12/5) x (-5/12) = -1
C1 correct conclusion: (5, 12) lies on both the circle and the line, and the radius to this point is perpendicular to the line, so (5, 12) must be the point of contact (cso)
Answer: (5, 12) is the point of contact (shown)

Question 19

(a) M1 finds the midpoint of AB as (5, 7) and recognises AB is horizontal, so its perpendicular bisector is vertical
(a) A1 x = 5 (cao)
(a) Answer: x = 5
(b) M1 finds the midpoint of BC as (9, 3) and recognises BC is vertical, so its perpendicular bisector is horizontal
(b) A1 y = 3 (cao)
(b) Answer: y = 3
(c) M1 recognises the centre lies on both perpendicular bisectors, so solves x = 5 and y = 3 simultaneously, ft their parts (a) and (b)
(c) A1 (5, 3) (cao)
(c) Answer: (5, 3)
(d) M1 uses the distance formula from their centre to A(1, 7) (or B or C), e.g. √(5-1)² + (3-7)²
(d) A1 radius = √32 (oe 4sqrt(2)), ft their part (c)
(d) A1 (x - 5)² + (y - 3)² = 32 (cao, ft)
(d) Answer: radius = √32 = 4sqrt(2); circle equation: (x - 5)² + (y - 3)² = 32

Perpendicular Lines and the equation of a tangent - Worksheets, Questions and Revision

8.9 Perpendicular Lines and the equation of a tangent

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Question 19