MEI C1 Study Resources Core1 Co-ordinate Geometry 1 Points and Straightlines REPACK DiRFiX

MEI Core 1 Coordinate Geometry Section 1: Points and straight lines Study Plan Background You will have met much of this work before at G.C.S.E.. The extra parts are extensions of what you already know. The formulae that are included need to be learnt so that they are at your fingertips and can be used in this topic and in later ones too. Work through the questions shown for each exercise as a minimum for practice checking the answers in the back of the textbook. Those questions with an asterisk beside them have worked solutions on the website to help you if you have problems.

Detailed work plan 1. Begin by reading pages 34 to 39 to remind yourself of the ideas in there and then try the questions specified below. There are some additional examples in the Notes and Examples. 2. You can also look at the Flash resources Gradient of a line, Distance between two points and Midpoint of two points. You may also find the Mathcentre videos Properties of straight line segments and The gradient of a straight line segment useful. 3. Exercise 2A Try all of Q1 taking care with signs particularly when dealing with gradients. Then attempt questions 2*, 4, 5*, 6, 7*, 10. 4. For extra practice try the interactive questions The gradient of a line, The gradient of a line between two points, Collinear points, The gradient of a perpendicular, The distance between two points and The midpoint between two points. 5. Read pages 42 to 44 to see all of the forms that the equation of a straight line can take. You will need to be able to sketch lines quickly no matter what form the equation takes in other parts of the course and practice will help. There are further examples in the Notes and Examples. Remember that lines parallel to the y-axis have infinite gradient and equation of the form x = k (k is a constant) whilst those parallel to the x axis have zero gradient and equation of the form y = k. 6. You can also try the Flash resources Equation of a line y = mx + c and Equation of a line ax + by + c = 0. You may also find the Mathcentre videos Linear functions and graphs and Equations of a straight line useful.

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MEI C1 Coordinate geometry Section 1 Study plan 7. Exercise 2B Try the odd numbered parts of both questions. 8. Familiarise yourself with the ways of obtaining the equation of a straight line depending on the information given. y – y1 = m(x – x1) is probably the most useful formula to learn. Remember when trying to decide whether lines have the same gradient or are perpendicular, the equation MUST be in the form y = mx + c. i.e. the coefficient of y must be 1. There are extra examples in the Notes and Examples. 9. You can also look at the Flash resources Equation of a line y – y1 = m(x – x1), Parallel lines and Perpendicular lines. 10. Exercise 2C Try the even parts of the first 3 questions then questions 5*, 7, 8*, 10*, 13. 11. For extra practice try the interactive questions The equation of a line between two points. 12. Read examples 1.33 and 1.34 in chapter 1 to remind yourself of the techniques for solving simultaneous equations, then read the examples on pages 55 and 56. There is an additional example in the Notes and Examples. 13. Also look at the Flash resource Intersection of two lines. 14. Exercise 2D Try questions 2, 4*, 6, 8*, 9, 10* 15. The Lines Activities include several different activities about the equations of lines and their properties.

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MEI Core 1 Coordinate Geometry Section 1: Points and straight lines Notes and Examples These notes contain sub-sections on:  Gradients, distances and mid-points  The equation of a straight line  The intersection of two lines

Gradients, distances and mid-points You will have met gradients before at GCSE. To revise finding the gradient of a line from a diagram, use the interactive questions The gradient of a line. Remember that lines which go “downhill” have negative gradients. To find the gradient of a straight line between two points  x1 , y1  and  x2 , y2  , use the formula y y gradient  2 1 . x2  x1 If two lines are parallel, they have the same gradient. If two lines with gradients m1 and m2 are perpendicular, then m1m2  1

Example 1 P is the point (-3, 7). Q is the point (5, 1). Calculate (i) the gradient of PQ (ii) the gradient of a line parallel to PQ (iii) the gradient of a line perpendicular to PQ. or vice versa: it will still give the same answer (WHY?)

Solution (i) Choose P as ( x1 , y1 ) and Q as ( x 2 , y 2 ). Gradient of PQ =

y 2  y1 1 7 6 3 = = = 4 x 2  x1 5  (3) 8 Notes: (1) Draw a sketch and check that your answer is sensible (e.g. has negative gradient). (2) Check that you get the same result when you choose Q as ( x1 , y1 ) and P as ( x2 , y2 ).

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MEI C1 Coordinate geom. Section 1 Notes & Examples (ii) When two lines are parallel their gradients are equal. ( m1 = m 2 ) 3 So the gradient of the line parallel to PQ is also  . 4 (iii) When two lines are perpendicular m1m2  1 . 3 So  m2  1 4 4  m2  3 4 The gradient of a line perpendicular to PQ is . 3 For more examples on gradient, look at the Flash resource Gradient of a line. You may also find the Mathcentre video The gradient of a straight line segment useful. For further practice in examples like the one above, try the interactive questions The gradient of a line between two points, Collinear points and The gradient of a perpendicular. The midpoint of a line joining two points  x1 , y1  and  x2 , y2  is given by

x x y y  Midpoint   1 2 , 1 2  2   2

 x2 , y2 

The x-coordinate of M is halfway between x1 and x2. The y-coordinate of M is halfway between y1 and y2.

M

 x1 , y1  The length of a line joining two points  x1 , y1  and  x2 , y2  can be found using Pythagoras’ Theorem. Length 

 x2  x1    y2  y1  2

2

 x2 , y2  y2  y1

 x1 , y1 

x2  x1

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MEI C1 Coordinate geom. Section 1 Notes & Examples Example 2 A is the point (2, -6). B is the point (-3, 4). Calculate (i) the midpoint of AB (ii) the length of AB.

or vice versa, it will still give the same answer (WHY?)

Choose A as ( x1 , y1 ) and B as ( x 2 , y 2 ). Solution (i)

Midpoint is i.e.

(ii)

 x1  x 2 y1  y 2  ,   2   2  2  (3)  6  4  ,   2 2    1  =  ,1  2 

The distance AB is given by

d  ( x1  x2 ) 2  ( y1  y2 ) 2  (2  (3)) 2  ((6)  4) 2  (5) 2  (10) 2  25  100  125

Note: The answer is often left like this if the square root is not exact. However since 125 = 25×5 then 125  25 5  5 5 is perhaps a simpler form.

To see more examples like these, try the Flash resources Distance between two points and Midpoint of two points. You may also find the Mathcentre video Properties of straight line segments useful. For further practice in examples like the one above, try the interactive questions The distance between two points and The midpoint between two points.

The equation of a straight line The equation of a straight line is often written in the form y = mx + c, where m is the gradient and c is the intercept with the y-axis.

Example 3 Find (i) the gradient and (ii) the y-intercept of the following straight-line equations. (a) 5 y  7 x  3 (b) 3x  8 y  7  0

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MEI C1 Coordinate geom. Section 1 Notes & Examples Solution (a) Rearrange the equation into the form y = mx + c. 5 y  7 x  3 becomes y  75 x  53 so m = 75 and c   53 (i) The gradient is

Note the minus sign

7 5

(ii) The y-intercept is  53 . (b) Rearrange the equation into the form y = mx + c. 3x  8 y  7  0 becomes 8 y  3x  7 giving y =  83 x+ 78 so m =  83 and c = 78 Note the minus sign

(i) The gradient is  83 (ii) The y-intercept is 78 .

Sometimes you may need to sketch the graph of a line. A sketch is a simple diagram showing the line in relation to the origin. It should also show the coordinates of the points where it cuts one or both axes. You can explore straight line graphs using the Flash resources Equation of a line y = mx + c and Equation of a line ax + by + c = 0. You may also find the Mathcentre video Equations of a straight line and Linear functions and graphs useful.

Example 4 Sketch the lines (a) 5 y  7 x  3

(b) 3x  8 y  7  0

Solution (a) From Example 3 you know that 5 y  7 x  3 has gradient

The gradient is positive, so the line slopes upwards from left to right. It’s also greater than 1 and so steeper than 45 degrees.

Sketch of 5 y  7 x  3

7 5

and y-intercept  53 .

This means the line goes through (0, -0.6) which is below the origin.

y

 53

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MEI C1 Coordinate geom. Section 1 Notes & Examples (b) From Example 3 you know that 3x  8 y  7  0 has gradient  83 and y-intercept 78 . The gradient is negative, so the line slopes downwards from left to right. It is also and less than 1 and so less steep than 45 degrees.

This means the line goes through (0, 0.875) which is above the origin.

Sketch of 3x  8 y  7  0 y 7 8

x

Sometimes you may need to find the equation of a line given certain information about it. If you are given the gradient and intercept, this is easy: you can simply use the form y = mx + c. However, more often you will be given the information in a different form, such as the gradient of the line and the coordinates of one point on the line (as in Example 5) or just the coordinates of two points on the line (as in Example 6). In such cases you can use the alternative form of the equation of a straight line. For a line with gradient m passing through the point  x1 , y1  , the equation of the line is given by y  y1  m  x  x1  .

Example 5 (i) Find the equation of the line with gradient 2 and passing through (3, -1). (ii) Find the equation of the line perpendicular to the line in (i) and passing through (3, -1). Solution (i) The equation of the line is y  y1  m  x  x1 

m = 2 and ( x1 , y1 ) is (3, -1)

 y   1  2  x  3  y 1  2x  6  y  2x  7

You should check that the point (3, -1) satisfies your line. If it doesn’t, you must have made a mistake!

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MEI C1 Coordinate geom. Section 1 Notes & Examples (ii)

For two perpendicular lines m1m2  1 , so the gradient of the new line is  12 . The equation of the line is y  y1  m  x  x1 

m   12 and

 y   1   12  x  3

 x1 , y1  is (3, -1)

 2 y  2  x  3  2 y  x  1  y   12 x  12

The final form of the equation can be written in various different ways: e.g. 2y = -x + 1 (This form has no fractions.) e.g. 2y + x = 1 (This has no fractions and avoids having a negative sign at the start of the right hand side.)

You can see more examples like this using the Flash resource Equation of a line y – y1 =m(x – x1). Also look at Parallel lines and Perpendicular lines.

In the next example, you are given the coordinates of two points on the line.

Example 6 P is the point (3, 8). Q is the point (-1, 5). Find the equation of PQ. Solution Choose P as ( x1 , y1 ) and Q as ( x 2 , y 2 ). Gradient of PQ =

Now use

One method is to find the gradient and then use this value and one of the points in

y  y1  m  x  x1 

y 2  y1 5  8 3 3 = = = x 2  x1  1  3 4 4

y  y1  m  x  x1 

 x  3  4  y  8   3  x  3  y 8 

3 4

 4 y  32  3 x  9  4 y  3 x  23

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You should check that P and Q satisfy your line.

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MEI C1 Coordinate geom. Section 1 Notes & Examples An alternative approach to the above examples is to put the formula for m into the straight line equation to obtain y y y  y1  2 1  x  x1  x2  x1 and then make the substitutions. This is equivalent to the first method, but does not involve calculating m separately first. For further practice in examples like the one above, try the interactive questions The equation of a line between two points.

The Lines Activities include a variety of activities which involve the equations of lines and their properties.

The intersection of two lines The point of intersection of two lines is found by solving the equations of the lines simultaneously. This can be done in a variety of ways. When both equations are given in the form y =… then equating the right hand sides is a good approach (see below). If both equations are not in this form, you can rearrange them into this form first, then apply the same method. Alternatively, you can use the elimination method (see Basic Algebra section 3) if the equations are in an appropriate form.

Example 7 Find the point of intersection of the lines y  3x  2 and y  5x  8 . Solution 3x  2  5 x  8

 2  2 x  8  6  2x  x3

Substitute into one of the equations to find y

Substituting x = 3 into y  3x  2 gives y  3  3  2  7 The point of intersection is (3, 7)

Check that (3, 7) satisfies the second equation.

You can see more examples like this using the Flash resource Intersection of two lines.

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MEI Core 1 Coordinate geometry Section 1: Points and straight lines Crucial points 1. Draw a diagram In most questions involving coordinate geometry, it is helpful to draw a sketch diagram. It does not need to be accurate, but it will help to give you a rough idea of the answer you might expect. 2. Ensure you can calculate the gradient of the line correctly. The gradient of a line, m, is given by change in y m change in x The gradient, m, of the line joining two points, ( x1 , y1 ) and ( x2 , y2 ) is given by y y m 2 1 x2  x1 Don’t get the gradient calculation upside-down! The gradient tells you by how much y changes when x increases by 1. 3. Make sure you can calculate the y-intercept of a straight-line graph. The y-intercept of a line is where it crosses the y-axis. It is the value of y when x = 0. 4. Make sure you understand how the standard straight-line equation works. An equation which can be written in the form y  mx  c represents a straight line. m is the gradient and c is the y-intercept. 5. Make sure you understand the conditions on the gradients of lines for the lines to be parallel or perpendicular. If two lines have gradients m1 and m2 then: 

the lines are parallel if m1  m2 .



the lines are perpendicular if m1m2  1 (i.e. if = m2  

1 ). m1

6. Make sure you understand and can remember how to calculate the distance between two points The distance, d, between two points,  x1 , y1  and  x2 , y2  is given by

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MEI C1 Coordinate geometry Section 1 Crucial points d  ( x1  x2 )2  ( y1  y2 )2 This is just from applying Pythagoras’s theorem 7. Make sure you understand and can remember how to calculate the midpoint of the line between two points. The coordinates of the midpoint, M, of the line joining  x1 , y1  and

 x2 , y2  are given by  x  x y  y2  M=  1 2, 1  2   2 8. Make sure you can calculate the equation of a straight line.  from the coordinates of two points on it.  from its gradient and the coordinates of a point on it.

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MEI Core 1 Coordinate geometry Section 1: Points and straight lines Exercise Do not use a calculator in this exercise. 1. (a) For the points A(3, 1) and B(7, 4) calculate (i) the gradient of AB (ii) the gradient of a line perpendicular to AB (iii) the midpoint of AB (iv) the distance AB (b) Repeat part (a) for the points A(-2, 9) and B(3, -1) 2. Given the points A(3, 1), B(6, y) and C(12, -2) find the value(s) of y for which (i) the line AB has gradient 2 (ii) the distance AB is 5 (iii) A, B and C are collinear (iv) AB is perpendicular to BC (v) the lengths AB and BC are equal 3. P is the point (2, 1), Q is (6, 9) and R is (10, 2). (i) Sketch the triangle PQR. (ii) Prove that triangle PQR is isosceles. (iii) Work out the area of triangle ABC. 4. The point E is (2, -1), F is (1, 3), G is (3, 5) and H is (4, 1). Show, by calculation that EFGH is a parallelogram. 5. Sketch the following lines. (i) y  x  3 (ii) y  2 x  1 (iv) 4 y  x  12 (v) 3 y  x  6  0

(iii) x  y  5 (vi) 5 y  15  2 x

6. Find the equations of the lines (a)-(e) in the diagram below. y

(a) 4 (d) (b)

(c)

4 x

-4

-4 (e)

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MEI C1 Coord. geom. Section 1 Exercise 7. Find the equations of the following lines. (i) parallel to y  4 x  1 and passing through (2, 3) (ii) perpendicular to y  2 x  7 and passing through (1, 2) (iii) parallel to 3 y  x  10 and passing through (4, -1) (iv) perpendicular to 3x  4 y  12 and passing through (-3, 0) (v) parallel to x  5 y  8  0 and passing through (-1, -6) 8. Find the equation of the line AB in each of the following cases. (i) A(1, 6), B(3, 2) (ii) A(8, -1), B(-2, 3) (iii) A(-5, 2), B(7, -4) (iv) A(-3, -5), B(5, 1) 9. A quadrilateral has vertices A(3, 5), B(9, 7), C(10, 4) and D(4, 2). (i) Sketch the quadrilateral. (ii) Find the equation of each of the quadrilateral’s sides. (iii) Use your equations to show that ABCD is a rectangle. 10. A triangle has vertices E(2, 5), F(4, 1) and G(-2, -3). (i) Find the midpoint of each side and hence find the equations of the three medians. (Medians are the lines from the midpoint of each side to the opposite vertex). (ii) Show that the point  43 ,1 lies on each median. 11. The sides of a triangle are formed by parts of the lines y  3x  11 , 3 y  x  3 and 7 y  x  37 . (i) Find the coordinates of the vertices of the triangle. (ii) Show that the triangle is right-angled. (iii) Work out the area of the triangle. 12. ABCD is a parallelogram. The equation of AB is y  4 x  3 and the equation of BC is y  2 x  1. (i) Find the coordinates of B. (ii) The coordinates of A are (3, 9). Find the equation of AD. (iii) The coordinates of C are (7, 15). Find the equation of CD. (iv) Find the coordinates of D.

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MEI Core 1 Coordinate geometry Section 1: Points and straight lines Solutions to Exercise 1. (a) (i) (ii)

Gradient of AB =

y 1 − y 2 1 − 4 −3 3 = = = . x 1 − x 2 3 − 7 −4 4

Gradient m of perpendicular line satisfies m × 43 = −1 4 so gradient of perpendicular line = − . 3

3+ 7 1 +4⎞ (iii) Midpoint of AB = ⎛⎜ , ⎟ = ( 5 , 2.5 ) 2 ⎠ ⎝ 2

(iv) Distance AB = (3 − 7)2 + (1 − 4)2

= 16 + 9 = 25 =5 (b) (i) (ii)

Gradient of AB =

y1 − y2 9 − ( −1) 10 = = = −2 . −2 − 3 −5 x1 − x2

Gradient m of perpendicular line satisfies m × −2 = −1 1 so gradient of perpendicular line = . 2

−2 + 3 9 + ( −1) ⎞ (iii) Midpoint of AB = ⎛⎜ , ⎟ = ( 0.5 , 4 ) 2 ⎝ 2 ⎠

(iv) Distance AB = ( −2 − 3)2 + ( 9 − ( −1))2 = 25 + 100 = 125 =5 5

y1 − y2 1 − y 1 − y = = −3 x1 − x2 3 − 6 1−y Gradient of AB = 2 ⇒ =2 −3 ⇒ 1 − y = −6 y=7

2. (i) Gradient of AB =

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MEI C1 Coord. Geom. Section 1 Exercise solns (ii) Distance AB is 5 (3 − 6)2 + (1 − y )2 = 5 9 + (1 − y )2 = 25 (1 − y )2 = 16 1 − y = ±4 y = 1 − 4 or 1 + 4

y = −3 or 5 (iii) If A, B and C are collinear, gradient of AB = gradient of AC. y 1 − y 2 1 − ( −2) 3 1 Gradient of AC = = = =− 3 − 12 −9 3 x1 − x2 1−y From (i), gradient of AB = −3 1−y 1 =− −3 3 1−y =1 y =0

(iv) If AB is perpendicular to BC, then grad AB × grad BC = -1 1−y From (i), gradient of AB = −3 y 1 − y 2 y − ( −2) y + 2 = = Gradient of BC = −6 x1 − x2 6 − 12 1−y y +2 × = −1 −3 −6 (1 − y )( y + 2) = −18 2 − y − y 2 = −1 8

y 2 + y − 20 = 0 ( y + 5 )( y − 4) = 0 y = −5 or y = 4 (v) Length AB = length BC

(3 − 6)2 + (1 − y )2 = (6 − 12)2 + ( y − ( −2))2 9 + (1 − y )2 = 36 + ( y + 2)2 1 − 2 y + y 2 = 27 + y 2 + 4 y + 4 0 = 6 y + 30 y = −5

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MEI C1 Coord. Geom. Section 1 Exercise solns 3. (i)

10

Q (6, 9)

y

5

R (10, 2)

P (2, 1)

x 2

4

6

8

10

12

(ii) PQ = (6 − 2)2 + ( 9 − 1)2 = 16 + 64 = 80 PR = (10 − 2)2 + (2 − 1)2 = 64 + 1 = 65 QR = (10 − 6)2 + (2 − 9)2 = 16 + 49 = 65 Since PR and QR are the same length, the triangle is isosceles. (iii) Take the base of the triangle as PQ. Let M be the midpoint of PQ 2 +6 1 + 9⎞ M = ⎛⎜ , ⎟ = ( 4, 5 ) 2 ⎠ ⎝ 2 Height of triangle is MR = (10 − 4)2 + (2 − 5 )2 = 36 + 9 = 45 Area of triangle = 21 × PQ × MR

=

1 2

80 45

=

1 2

16 × 5

9× 5

= 21 × 4 5 × 3 5 = 6× 5 = 30 3 − ( −1) 4 = = −4 1−2 −1 5 −3 2 = =1 Gradient of FG = 3−1 2 1 − 5 −4 Gradient of GH = = = −4 4−3 1 1 − ( −1) 2 = =1 Gradient of EH = 4−2 2

4. Gradient of EF =

EF is parallel to GH and FG is parallel to EH so EFGH is a parallelogram.

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MEI C1 Coord. Geom. Section 1 Exercise solns 5. (i)

y = x +3 3

Gradient = 1 When x = 0, y = 3 When y = 0, x + 3 = 0 ⇒ x = −3

(ii) y = 2 x − 1 Gradient = 2 When x = 0, y = -1 When y = 0, 2 x − 1 = 0 ⇒ x =

-3

1 2

1 2

-1

(iii) x + y = 5 y = −x + 5 Gradient = -1 When x = 0, y = 5 When y = 0, x = 5

5 5

(iv) 4 y = x + 12

y = 41 x + 3 Gradient = 41 When x = 0, y = 3 When y = 0, x + 12 = 0 ⇒ x = −12

3 -12

(v) 3 y + x + 6 = 0 3y = − x − 6

y = − 31 x − 2 Gradient = − 31 When x = 0, y = -2 When y = 0, x + 6 = 0 ⇒ x = −6

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MEI C1 Coord. Geom. Section 1 Exercise solns (vi) 5 y = 15 − 2 x

y = 3 − 52 x Gradient = − 52 When x = 0, 5 y = 15 ⇒ y = 3 When y = 0, 15 − 2 x = 0 ⇒ x = 7.5

3 7.5

6. (a) Gradient = 1, y-in tercept = 2 Equation of line is y = x + 2 (b) Gradient = 21 , y-intercept = -1 Equation of line is y = 21 x − 1 or 2 y = x − 2 (c) Gradient = − 21 , y-intercept = -2 Equation of line is y = − 21 x − 2 or 2 y + x + 4 = 0 (d) Gradient = − 41 ,y- intercept = 3 Equation of line is y = − 41 x + 3 or 4 y + x = 12 (e) Gradient = − 83 , passes through (-1, 4) Equation of line is y − 4 = − 83 ( x − ( −1))

3( y − 4) = −8( x + 1) 3 y − 12 = −8 x − 8 3y + 8 x = 4 7. (i) Gradient of y = 4 x − 1 is 4 Gradient of parallel line = 4 Equation of line is y − 3 = 4( x − 2) y − 3 = 4x − 8 y = 4x − 5 (ii) Gradient of y = 2 x + 7 is 2 Gradient of perpendicular line is − 21

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MEI C1 Coord. Geom. Section 1 Exercise solns Equation of line is y − 2 = − 21 ( x − 1) 2( y − 2) = −( x − 1) 2y − 4 = −x + 1 2y + x = 5 (iii) 3 y + x = 10 ⇒ y = − 31 x +

10 3

Gradient is − 31 Gradient of parallel line is − 31 Equation of line is y − ( −1) = − 31 ( x − 4) 3( y + 1) = −( x − 4) 3y + 3 = − x + 4 3y + x = 1 (iv) 3 x + 4 y = 12 ⇒ y = − 43 x + 3 Gradient is − 43 Gradient of perpendicular line is 43 Equation of line is y − 0 = 43 ( x − ( −3)) 3 y = 4( x + 3) 3 y = 4 x + 12 (v) x + 5 y + 8 = 0 ⇒ y = − 51 x − 58 Gradient is − 51 Gradient of parallel line is − 51 Equation of line is y − ( −6) = − 51 ( x − ( −1)) 5( y + 6) = −( x + 1) 5 y + 30 = − x − 1 5 y + x + 31 = 0

y1 − y2 6 − 2 4 = = = −2 x 1 − x 2 1 − 3 −2 Equation of AB is y − 6 = −2( x − 1) y − 6 = −2 x + 2 y + 2x = 8

8. (i) Gradient of AB =

(ii) Gradient of AB =

y1 − y2 −1 − 3 −4 2 = =− = 5 x 1 − x 2 8 − ( −2) 10

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MEI C1 Coord. Geom. Section 1 Exercise solns Equation of AB is y − ( −1) = − 52 ( x − 8) 5( y + 1) = −2( x − 8) 5 y + 5 = −2 x + 16 5 y + 2 x = 11

y 1 − y 2 2 − ( −4) 6 1 = = =− −5 − 7 −12 x1 − x2 2 1 Equation of AB is y − 2 = − 2 ( x − ( −5 )) 2( y − 2) = −( x + 5 ) 2y − 4 = −x − 5 2y + x + 1 = 0

(iii) Gradient of AB =

y 1 − y 2 −5 − 1 − 6 3 = = = x 1 − x 2 − 3 − 5 −8 4 Equation of AB is y − ( −5 ) = 43 ( x − ( −3)) 4( y + 5 ) = 3( x + 3) 4 y + 20 = 3 x + 9 4 y = 3 x − 11

(iv) Gradient of AB =

9. (i)

8 7 6

B (9, 7)

y

A (3, 5)

5

C (10, 4)

4 3 2

D (4, 2)

1

x 2

4

6

8

10

12

−1 −2

7−5 2 1 = = 9−3 6 3 Equation of AB is y − 5 = 31 ( x − 3)

(ii) Gradient of AB =

3( y − 5 ) = x − 3 3 y − 15 = x − 3 3 y = x + 12 Gradient of BC =

4− 7 −3 = = −3 10 − 9 1

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MEI C1 Coord. Geom. Section 1 Exercise solns Equation of BC is y − 7 = −3( x − 9) y − 7 = −3 x + 27 y + 3 x = 34 2 − 4 −2 1 = = 4 − 10 −6 3 Equation of CD is y − 4 = 31 ( x − 10)

Gradient of CD =

3( y − 4) = x − 10 3 y − 12 = x − 10 3y = x + 2 2 − 5 −3 = = −3 4−3 1 Equation of AD is y − 5 = −3( x − 3) y − 5 = −3 x + 9 y + 3 x = 14

Gradient of AD =

AB and CD are parallel, and BC and AD are parallel. The gradients of AB and CD are 31 , and the gradients of BC and AD are -3. Since 31 × −3 = −1 , AB and CD are perpendicular to BC and AD so ABCD is a rectangle. 2+4 5 +1⎞ 10. (i) Midpoint of EF = ⎛⎜ , ⎟ = (3 , 3) 2 ⎠ ⎝ 2 4 + ( −2) 1 + ( −3) ⎞ Midpoint of FG = ⎛⎜ , ⎟ = (1 , −1) 2 2 ⎝ ⎠ 2 + ( −2) 5 + ( −3) ⎞ Midpoint of EG = ⎛⎜ , ⎟ = ( 0, 1) 2 2 ⎝ ⎠

Median from midpoint of EF (3, 3) to G (-2, -3) −3 − 3 −6 6 = = Gradient of median = −2 − 3 −5 5 Equation of median is y − 3 = 56 ( x − 3) 5( y − 3) = 6( x − 3) 5 y − 15 = 6 x − 18 5 y = 6x − 3 Median from midpoint of FG (1, -1) to E (2, 5) 5 − ( −1) 6 Gradient of median = = =6 2−1 1

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MEI C1 Coord. Geom. Section 1 Exercise solns Equation of median is y − ( −1) = 6( x − 1) y + 1 = 6x − 6 y = 6x − 7 Median from midpoint of EG (0, 1) to F (4, 1) 1−1 0 Gradient of median = = =0 4−0 4 Equation of median is y = 1 (ii) Equation of first median is 5 y = 6 x − 3 Substituting x = so

4 3

gives 5 y = 6 × 43 − 3 = 8 − 3 = 5 y=1

( 43 , 1 ) lies on the median.

Equation of second median is y = 6 x − 7 Substituting x = so

4 3

gives y = 6 × 43 − 7 = 8 − 7 = 1

( 43 , 1 ) lies on the median.

Equation of third median is y = 1 , so

( 43 , 1 ) lies on the median.

11. (i) Let the triangle be ABC. Let A be the intersection point of y + 3 x = 11 and 3 y = x + 3 . y + 3 x = 11 ⇒ y = 11 − 3 x Substituting into 3 y = x + 3 gives 3(11 − 3 x ) = x + 3 33 − 9 x = x + 3

When x = 3, y = 11 − 3 × 3 = 2 The coordinates of A are (3, 2).

30 = 10 x x =3

Let B be the intersection point of 3 y = x + 3 and 7 y + x = 37 3y = x + 3 ⇒ x = 3y − 3 Substituting into 7 y + x = 37 gives 7 y + 3 y − 3 = 37 10 y = 40

y =4 When y = 4, x = 3 × 4 − 3 = 9 The coordinates of B are (9, 4). Let C be the intersection point of 7 y + x = 37 and y + 3 x = 11 y + 3 x = 11 ⇒ y = 11 − 3 x

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MEI C1 Coord. Geom. Section 1 Exercise solns Substituting into 7 y + x = 37 gives 7(11 − 3 x ) + x = 37 77 − 21 x + x = 37 40 = 20 x x =2

When x = 2, y = 11 − 3 × 2 = 5 The coordinates of C are (2, 5).

The vertices of the triangle are (3, 2), (9, 4) and (2, 5). (ii) AB is the line y + 3 x = 11 ⇒ y = 11 − 3 x so the gradient of AB is -3. BC is the line 3 y = x + 3 ⇒ y = 31 x + 1 so the gradient of BC is 31 . Gradient of AB × gradient of AC = −3 × 31 = −1 so AB and AC are perpendicular, and therefore the triangle is rightangled. (iii) 6

y

C (2, 5)

5

B (9, 4)

4 3 2

A (3, 2)

1

x 1

2

3

4

5

6

7

8

9

10

AB = (3 − 9)2 + (2 − 4)2 = 36 + 4 = 40 AC = (3 − 2)2 + (2 − 5 )2 = 1 + 9 = 10 Area of triangle = 21 × AB × AC =

1 2

40 10

=

1 2

4 10 10

= 21 × 2 × 10 = 10 12. (i) B is the intersection point of y = 4 x − 3 and y = 2 x + 1 . 4x − 3 = 2 x + 1 2x = 4 x =2 When x = 2, y = 4 × 2 − 3 = 5 The coordinates of B are (2, 5).

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MEI C1 Coord. Geom. Section 1 Exercise solns (ii) AD is parallel to BC, so AD has gradient 2. AD passes through the point (3, 9). Equation of AD is y − 9 = 2( x − 3) y − 9 = 2x −6 y = 2x + 3

y = 4x – 3

(iii) CD is parallel to AB, so CD has gradient 4. CD passes through the point (7, 15). Equation of CD is y − 15 = 4( x − 7) y − 15 = 4 x − 28 y = 4 x − 13

B

D C

y = 2x + 1

A

(iv) D is the intersection point of AD and CD. 2 x + 3 = 4 x − 13 16 = 2 x x =8 When x = 8, y = 2 × 8 + 3 = 19 The coordinates of D are (8, 19).

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Linear Functions Student sheets Match the Equations 1 Match the equations to the line (3 for each), and be prepared to have to give a reason for your choice!

(2, 2)

A

(0, -4)

y = 3x – 4

y + 3x + 4 = 0

y + 4 = 3x

2y = 3x + 1

4y = 6x + 2

3y + 9x = –12

2y – 3x = 1

2y – 6x + 8 = 0

C (-2, 2)

(0, -4)

y = – 3x – 4

(3, 5)

B (1, 2)

© Susan Wall, MEI, 2004

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Linear Functions Student sheets Match the equations 2 Label the lines with these equations:

y  4x  3

y  3  2x

y  2x  3

y  4  3x

y  3x  4  0

y  3x

y  3x  4

y  3  4x

y  3x  0 y  3x  4  0

© Susan Wall, MEI, 2004

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Linear Functions Student sheets Pairs

y = 4x + 4

4y = x + 3

y = 8x – 3

y + 4x + 6 = 0

3y = 2x – 8

y + 6x = 11

y + 8x = 6

2y + 8 = 3x

2y + x = 4

2y = 8x + 3

y = 6x – 4

y+x+8=0

© Susan Wall, MEI, 2004

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Linear Functions Student sheets  Find two lines from the “Pairs” sheet that fit each of the properties below.  You will have two lines left over. Find a property for this pair of lines.  Add a line of your own to each set where possible.

These lines are parallel.

These lines are perpendicular.

These lines have the same y intercept.

These lines have the same x intercept.

These lines both go through the point (1, 5).

These lines ….

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Linear Functions Student sheets Jigsaw Problem Cut out the stages of the solution and sort them into the correct order. Add comments to explain the solution. Question:

Three points have co-ordinates A (1, 7), B (7, 5) and C (0, -2). Find: (a) The equation of the perpendicular bisector of AB. (b) The equation of the line BC. (c) The point of intersection of these two lines.

-------------------------------------------------------------y=x-2 -------------------------------------------------------------4 = 2x  x = 2 -------------------------------------------------------------y = 3x + c -------------------------------------------------------------y = 3x - 6 -------------------------------------------------------------Point is (2, 0) -------------------------------------------------------------Gradient = -

2 1 =6 3

-------------------------------------------------------------6 = (3  4) + c  c = - 6 -------------------------------------------------------------x - 2 = 3x - 6 --------------------------------------------------------------2=c -------------------------------------------------------------Point is (4, 6) -------------------------------------------------------------Gradient is 3 -------------------------------------------------------------y=x+c -------------------------------------------------------------Gradient is 1 -------------------------------------------------------------1 Gradient = 1 3

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Linear Functions Student sheets Intersection of Lines Find where the following pairs of lines intersect:

a)

y  2x  3 y  4x  1

d)

4 x  3 y  2 y  3x  1

b)

y  4  2x y  x 1

c)

y  2x  3 3x  2 y  1

Solutions – please mark and correct:

a)

2x  3  4x  1 2x  4  4x 4  2x x2

b)

4  2x  x  1 2x  x  5 x5 y6

c)

3x  4 x  3  1 7x  3  1 7x  4 4 x 7 8 5 y  3 7 7

d)

4 x  3(3 x  1)  2 4 x  9 x  3  2 5 x  3  2 5 x  1 1 5 3 8 y   1  5 5

x

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Linear Functions Student sheets Properties of Straight Lines Match each of the properties below to one of the equations:

y = 3 – 2x 3y + 2x = 12

2y = x + 6 y = 2x – 3

y = 3x + 2 y = 4x

y intercept is 2

Parallel to y = 2x + 7

Passes through (0, 0)

Passes through (1, 5)

Passes through (0, 4)

y intercept is –3

x intercept is



2 3

y intercept is 4

Gradient is –2

Gradient is 2

Parallel to y = 3x – 1

Perpendicular to y = 3 – 2x

Perpendicular to y = 5



x 4

Passes through (3, 2)

Gradient is 3

Perpendicular to y =

x intercept is –6

x intercept is 6

© Susan Wall, MEI, 2004

3 x 2

+1

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MEI Core 1 Coordinate Geometry Points and straight lines Multiple Choice Test Do not use a calculator for this test 1. Which of the following points does not lie on the line 2y + 5x – 4 = 0? (a) (0.8, 0) (c) (0, 2) (e) I don’t know

(b) (1, -0.5) (d) (2, 3)

2. Here are four straight-line equations. 1 3

3y = 4x + 5 4y + 3x = 7

2 4

4y = 3x – 1 4x + 3y = 2

Which one of the following statements is true? (a) Lines 1 and 2 are perpendicular (c) Lines 2 and 4 are perpendicular I don’t know

(b) Lines 1 and 4 are parallel (d) Lines 2 and 3 are parallel

3. A straight line has equation 10y = 3x + 15. Which of the following is true? (a) The gradient is 0.3 and the y-intercept is 1.5 (b) The gradient is 3 and the y-intercept is 15 (c) The gradient is 15 and the y-intercept is 3 (d) The gradient is 1.5 and the y-intercept is 0.3 (e) I don’t know

4.

The diagram shows a sketch of one of the following lines. Which one? (a) y – x + 1 = 0 (c) y = x + 1 (e) I don’t know

(b) y + x = 1 (d) y + x + 1 = 0

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MEI C1 Coord. geometry Section 1 MC test solutions 5. P is the point (2, 7). Q is the point (6, -3). What is the gradient of PQ? (a) 0.4 (c) 2.5 (e) I don’t know

(b) -0.4 (d) -2.5

6. P is the point (4, -2). Q is the point (-3, -5). What is the length PQ? (a)

50

(c) 40 (e) I don’t know

(b)

98

(d)

58

7. P is the point (3, 5). Q is the point (-1, 9). R is the midpoint of PQ. On which one of the following lines does R lie? (a) y = x + 6 (c) y = x – 6 (e) I don’t know

(b) y = x + 8 (d) y = x – 8

8. A straight line has a gradient of –2 and passes through the point (4, 1). What is its equation? (a) y + 2x = 6 (c) y + 2x – 9 = 0 (e) I don’t know

(b) y = 2x – 6 (d) 2y = x – 2

9. The lines y = 5x – 3 and y = 2x + 9 intersect at P. What are the coordinates of P? (a) (2, 7) (c) (4, 17) (e) I don’t know

(b) (2, 13) (d) (-4, -23)

10. A is the point (1, 5), B is the point (4, 7) and C is the point (5, 2). Triangle ABC is (a) right-angled (c) equilateral (e) I don’t know

(b) scalene with no right angle (d) isosceles

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