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Figure 5.1 shows some scaffolding in which some of the horizontal pieces are 2 m long and others are 1 m. All the vertical pieces are 2 m.
➜ An ant crawls along the scaffolding from point P to point Q, travelling either horizontally or vertically. How far does the ant crawl?
➜ A mouse also goes from point P to point Q, travelling either horizontally or along one of the sloping pieces. How far does the mouse travel?
➜ A bee fl ies directly from point P to point Q. How far does the bee fl y?
A place for everything,
and everything in its
place.
samuel smiles (1812–1904)
Coordinate geometry
Figure 5.
P
Q
Working with coordinates
1 Working with coordinates
Coordinates are a means of describing a position relative to a fixed point, or origin. In two dimensions you need two pieces of information; in three dimensions you need three pieces of information. In the Cartesian system (named after René Descartes), position is given in perpendicular directions: x , y in two dimensions; x , y , z in three dimensions. This chapter concentrates exclusively on two dimensions.
The midpoint and length of a line segment When you know the coordinates of two points you can work out the midpoint and length of the line segment which connects them.
ACTIVITY 5.
Find (i) the coordinates of the midpoint, M (ii) the length AB.
Draw a right-angled triangle with AB as the hypotenuse and use Pythagoras’ theorem.
y
O x
A (2, 1)
B (8, 5)
M
Figure 5.
You can generalise these methods to find the midpoint and length of any line segment AB. Let A be the point ( x (^) 1 , y 1 )and B the point ( x (^) 2 , y 2 ). (i) Find the midpoint of AB. The midpoint of two values is the mean of those values. The mean of the x coordinates is x^ x 2
1 +^2.
The mean of the y coordinates is y^ y 2
1 +^2.
So the coordinates of the midpoint are x x y y 2 ,^2
1 +^2 1 + 2
(ii) Find the length of AB. First find the lengths of AC and AB: x^ x y y
AC
BC
2 1 2 1
By Pythagoras’ theorem: x x y y
AB AC BC
2 2 2
2 1
2 2 1
2
So the length AB is ( x (^) 2 − x (^) 1 ) 2 + ( y (^) 2 − y 1 )^2
y
x
B ( x 2 , y 2 )
C ( x 2 , y 1 )
A ( x 1 , y 1 )
O
C has the same x coordinate as B… …and the same y coordinate as A.
Figure 5.
TeCHnology
When working through this chapter, you may wish to use a graphical calculator or graphing software to check your answers where appropriate.
Working with coordinates
Example 5.1 (^) A and B are the points (2, 5) and (6, 3) respectively (see Figure 5.6).
Find: (i) the gradient of AB (ii) the length of AB (iii) the midpoint of AB (iv) the gradient of the line perpendicular to AB.
Solution
1 2
(ii) Length AB = ( x (^) B − x (^) A) 2 + ( y (^) B − y A)^2
(6 2) (3 5) 16 4 20
= − 2 + −^2
(iii) Midpoint =
x x y y 2 ,^2
A +^ B A + B
= (^) ( )
(iv) Gradient of AB: m AB = − (^12)
So gradient of perpendicular to AB is 2.
y
O x
B (6, 3)
A (2, 5)
Figure 5.
(i) Gradient =
m (^) −
y y AB x x A B A B
Gradient is difference in y coordinates divided by difference in x coordinates. It doesn’t matter which point you use first, as long as you are consistent!
The gradient of the line perpendicular to AB is the negative reciprocal of m AB.
Draw a diagram to help you.
Check: −− 12 × 2 = −1 ✓
5
Chapter 5 Coordinate geometry
Example 5.2 (^) The points P(2, 7), Q(3, 2) and R(0, 5) form a triangle.
(i) Use gradients to show that RP and RQ are perpendicular. (ii) Use Pythagoras’ theorem to show that PQR is right-angled.
Solution
(i) Show that the gradients satisfy m 1 m 2 = −
Gradient of RP = 27 −− 05 = 1
Gradient of RQ = 23 −− 05 = − 1 ⇒ product of gradients = 1 × (–1) = – ⇒ sides RP and RQ are at right angles. (ii) Pythagoras’ theorem states that for a right-angled triangle with hypotenuse of length a and other sides of lengths b and c , a^2 = b^2 + c^2. Conversely, when a^2 = b^2 + c^2 for a triangle with sides of lengths a , b and c , then the triangle is right-angled and the side of length a is the hypotenuse. length 2 = ( x (^) 2 − x (^) 1 ) 2 + ( y (^) 2 − y 1 )^2 PQ 2 = (3 – 2)^2 + (2 – 7)^2 = 1 + 25 = 26 RP 2 = (2 – 0)^2 + (7 – 5)^2 = 4 + 4 = 8 RQ 2 = (3 – 0)^2 + (2 – 5)^2 = 9 + 9 = 18 Since 26 = 8 + 18, PQ^2 = RP^2 + RQ^2 ⇒ sides RP and RQ are at right angles.
y
O x
P (2, 7)
Q (3, 2)
R (0, 5)
Figure 5.
PQ is the hypotenuse since RP and RQ are perpendicular.
Always start by drawing a diagram.
5
Chapter 5 Coordinate geometry
Example 5.3 (^) (i) Sketch the lines (a) y = x – 1 and (b) 3 x + 4 y = 24 on the same axes.
(ii) Are these lines perpendicular?
Solution (i)
(a)
Substituting y = 0 gives x = 1, so the line also passes through (1, 0).
(b)
Substituting x = 0 gives 4 y = 24 ⇒ y = 6 substituting y = 0 gives 3 x = 24 ⇒ x = 8. So the line passes through (0, 6) and (8, 0). (Figure 5.10 overleaf)
Usually it is easiest to find where the line cuts the x and y axes.
To draw a line you need to find the coordinates of two points on it.
The line y = x – 1 passes through the point (0, –1).
The line is already in the form y = mx + c.
Find two points on the line 3 x + 4 y = 24.
Set x = 0 and find y to give the y -intercept. Then set y = 0 and find x to give the x -intercept.
2 The equation of a straight line
Drawing a line, given its equation There are several standard forms for the equation of a straight line, as shown in Figure 5.9.
Figure 5.
y
(3, 0)^ x
x = 3
O
y
x
(0, 2) (^) y = 2
O
y
(3, 0) x
x = 3
O
y
x
(0, 2) (^) y = 2
O
(a) Equations of the form x (^) = a (b) Equations of the form y = b
(c) Equations of the form y = mx (d) Equations of the form y = mx + c y
x
y = –4 x (^) y = 1 – 2 x
O
y
x
(0, 2)
(3, 0)
2 x + 3 y – 6 = 0
O
y
x
(0, 1)
(1, 0)
(0, –1)
(3, 0)
y = x – 1
O y = – –^13 x + 1
(e) Equations of the form px + qy + r = 0
y
x
y = –4 x (^) y = 1 – 2 x
O
y
x
(0, 2)
(3, 0)
2 x + 3 y – 6 = 0
O
y
x
(0, 1)
(1, 0)
(0, –1)
(3, 0)
y = x – 1
O y = –^1 – 3 x + 1
y
x
y = –4 x (^) y = 1 – 2 x
O
y
x
(0, 2)
(3, 0)
2 x + 3 y – 6 = 0
O
y
x
(0, 1)
(1, 0)
(0, –1)
(3, 0)
y = x – 1
O y = – –^13 x + 1
Each point on the line has an x coordinate of 3.
Each point on the line has a y coordinate of 2.
This is often a tidier way of writing the equation.
These lines have gradient m and cross the y axis at point (0, c ).
These are lines through the origin, with gradient m.
All such lines are parallel to the y axis.
All such lines are parallel to the x axis.
y
x
(8, 0)
(0, –1)
3 x + 4 y = 24
y = x – 1
(1, 0)
(0, 6)
0
–1 1 2 3 4 5 6 7 8
1
2
3
4
5
6
Figure 5. (ii) The lines look almost perpendicular but you need to use the gradient of each line to check.
Therefore the lines are not perpendicular as 1 × (^) ( −^34 )≠ −1.
Gradient of y = x − 1 is 1.
Gradient of 3 x + 4 y = 24 is − 43.
Rearrange the equation to make y the subject so you can fi nd the gradient. 4 y = −3 x + 24 y = − 34 x + 6
Finding the equation of a line To fi nd the equation of a line, you need to think about what information you are given. (i) Given the gradient, m , and the coordinates ( x 1 , y 1 ) of one point on the line Take a general point ( x , y ) on the line, as shown in Figure 5.11. y
x
( x 1 , y 1 )
( x , y )
O
Figure 5. The gradient, m , of the line joining ( x 1 , y 1 ) to ( x , y ) is given by
= −−
⇒ − = −
m yx^ yx
y y m x ( x )
1 1 ⇒ 1 1
m
y y x x y y m x ( x )
1 1 1 1 For example, the equation of the line with gradient 2 that passes through the point (3, −1) can be written as y − ( 1)− = 2( x −3) which can be simplified to y = 2 x − 7.
This is a very useful form of the equation of a straight line.
The equation of a straight line
y − y (^) 1 = m x ( − x 1 )
Warning
When you draw two perpendicular lines on a diagram, they will be at right angles if, and only if, both axes are to the same scale.
Example 5.4 (^) Find the equation of the line perpendicular to 4y + x = 12 which passes
through the point P(2, −5).
Solution First rearrange 4y + x = 12 into the form y = mx + c to find the gradient. 4 y = −x + 12
y = − 14 x+ 3
So the gradient is (^) −^14 The negative reciprocal of − 41 is 4.
So the gradient of a line perpendicular to y^ = −^14 x+^3 is 4. Using y − y 1 = m(x − x 1 ) when m = 4 and (x 1 , y 1 ) is (2, −5) ⇒ y − (−5) = 4(x − 2) ⇒ y + 5 = 4x − 8 ⇒ y = 4x − 13
For perpendicular gradients m 1 m 2 = − So m 2 = −m^11
Check: (^) − 14 × 4 = −1✓
Straight lines can be used to model real-life situations. Often simplifying assumptions need to be made so that a linear model is appropriate.
Solution (i) The graph passes through the points (0, 9) and (140, 23). y
O x
(140, 23)
(0, 9)
diameter (mm)
distance from tip (cm) Figure 5.
Example 5.5 (^) The diameter of a snooker cue
varies uniformly from 9 mm to 23 mm over its length of 140 cm. (i) Sketch the graph of diameter (y mm) against distance (x cm) from the tip. (ii) Find the equation of the line. (iii) Use the equation to find the distance from the tip at which the diameter is 15 mm.
Varying uniformly means that the graph of diameter against distance from the tip is a straight line.
The equation of a straight line
5
Chapter 5 Coordinate geometry
(ii) Gradient =
y y x x
2 1 2 1
Using the form y = mx + c , the equation of the line is y = 0.1 x + 9. (iii) Substituting y = 15 into the equation gives 15 = 0.1 x + 9 0.1 x = 6 x = 60 ⇒ The diameter is 15 mm at a point 60 cm from the tip.
Discussion points
➜ Which of these situations in Figure 5.16 could be modelled by a straight line? ➜ For each straight line model, what information is given by the gradient of the line? ➜ What assumptions do you need to make so that a linear model is appropriate? ➜ How reasonable are your assumptions?
Figure 5.
Interest earned on savings in a bank account against time
Height of ball dropped from a cliff against time
Profit of ice cream seller against number of sales Tax paid against earnings Cost of apples against mass of apples
Value of car against age of car
Mass of candle versus length of time it is burning
Distance travelled by a car against time
Mass of gold bars against volume of gold bars Population of birds on an island against time
Mobile phone bill against number of texts sent
Length of spring against mass of weights attached
① Sketch the following lines:
(i) y = − (iii) y = −2 x (v) y = 2 x + 5 (vii) 2 x − y = 5
(ii) x = 2 (iv) y = x + 2 (vi) y = 5 − 2 x (viii) y + 2 x + 5 = 0
② Find the equations of the lines (i)–(v) in Figure 5.17.
–4 –2 0 2 4 6 8
2
4
6
x
y (iii) (ii)
(i)
(v)
(iv)
Figure 5.
③ Find the equations of the lines (i) parallel to y = 3 x − 2 and passing through (0, 0) (ii) parallel to y = 3 x and passing through (2, 5) (iii) parallel to 2 x + y − 3 = 0 and passing through (−2, 5) (iv) parallel to 3 x − y − 2 = 0 and passing through (5, −2) (v) parallel to x + 2 y = 3 and passing through (−2, −5). ④ Find the equations of the lines (i) perpendicular to y = 3 x and passing through (0, 0) (ii) perpendicular to y = 2 x + 3 and passing through (4, 3) (iii) perpendicular to 2 x + y = 4 and passing through (4, −3)
Exercise 5.
5
Chapter 5 Coordinate geometry
⑫ A spring has an unstretched length of 10 cm. When it is hung with a load of 80 g attached, the stretched length is 28 cm. Assuming that the extension of the spring is proportional to the load: (i) draw a graph of extension E against load L and find its equation (ii) find the extension caused by a load of 48 g (iii) find the load required to extend the spring to a length of 20 cm. This particular spring passes its elastic limit when it is stretched to four times its original length. (This means that if it is stretched more than that it will not return to its original length.)
(iv) Find the load which would cause this to happen. ⑬
( a , 0) O
y
x
(0, b )
Figure 5. Show that the equation of the line in Figure 5.19 can be written x a
y
PS
PS
3 The intersection of two lines
The intersection of any two curves (or lines) can be found by solving their equations simultaneously. In the case of two distinct lines, there are two possibilities: (i) they are parallel, or (ii) they intersect at a single point. You often need to find where a pair of lines intersect in order to solve problems.
Example 5.
Solution You need to solve the equations y = 5 x – 13 ① and 2 y + 3 x = 0 ② simultaneously. Substitute equation ① into ② : 2(5 x – 13) + 3 x = 0 10 x – 26 + 3 x = 0 13 x – 26 = 0 13 x = 26 x = 2 Substitute x = 2 into equation ① to find y. y = 5 × 2 − 13 y = − So the coordinates of P are (2, −3).
Multiply out the brackets.
Simplify
Don’t forget to find the y coordinate.
The lines y = 5 x − 13 and 2 y + 3 x = 0 intersect at the point P. Find the coordinates of P.
Discussion point
➜ The line l has equation 2 x − y = 4 and the line m has equation y = 2 x − 3. What can you say about the intersection of these two lines?
① Find the coordinates of the point of intersection of the following pairs of lines. (i) y = 2 x + 3 and y = 6 x + 1 (ii) y = 2 − 3 x and 2 y + x = 14 (iii) 3 x + 2 y = 4 and 5 x − 4 y = 3 ② (i) Find the coordinates of the points where the following pairs of lines intersect. (a) y = 2 x − 4 and 2 y = 7 − x (b) y = 2 x + 1 and 2 y = 7 − x The lines form three sides of a square. (ii) Find the equation of the fourth side of the square. (iii) Find the area of the square. ③ (i) Find the vertices of the triangle ABC whose sides are given by the lines AB: x − 2 y = − BC: 7 x + 6 y = 53 and AC: 9 x + 2 y = 11. (ii) Show that the triangle is isosceles. ④ A(0, 1), B(1, 4), C(4, 3) and D(3, 0) are the vertices of a quadrilateral ABCD. (i) Find the equations of the diagonals AC and BD. (ii) Show that the diagonals AC and BD bisect each other at right angles. (iii) Find the lengths of AC and BD. (iv) What type of quadrilateral is ABCD? ⑤ The line y = 5 x − 2 crosses the x axis at A. The line y = 2 x + 4 crosses the x axis at B. The two lines intersect at P. (i) Find the coordinates of A and B. (ii) Find coordinates of the point of intersection, P. (iii) Find the exact area of the triangle ABP. ⑥ Triangle ABC has an angle of 90° at B. Point A is on the y axis, AB is part of the line x – 2 y + 8 = 0, and C is the point (6, 2). (i) Sketch the triangle. (ii) Find the equations of the lines AC and BC.
(iii) Find the lengths of AB and BC and hence find the area of the triangle. (iv) Using your answer to (iii), find the length of the perpendicular from B to AC. ⑦ Two rival taxi firms have the following fare structures: Firm A: fixed charge of £1 plus 40p per kilometre; Firm B: 60p per kilometre, no fixed charge. (i) Sketch the graph of price (vertical axis) against distance travelled (horizontal axis) for each firm (on the same axes). (ii) Find the equation of each line. (iii) Find the distance for which both firms charge the same amount. (iv) Which firm would you use for a distance of 6 km? ⑧ Two sides of a parallelogram are formed by parts of the lines 2 x − y = − and x − 2 y = −9. (i) Show these two lines on a graph. (ii) Find the coordinates of the vertex where they intersect. Another vertex of the parallelogram is the point (2, 1). (iii) Find the equations of the other two sides of the parallelogram. (iv) Find the coordinates of the other two vertices. ⑨ The line with equation 5 x + y = 20 meets the x axis at A and the line with equation x + 2 y = 22 meets the y axis at B.The two lines intersect at a point C. (i) Sketch the two lines on the same diagram. (ii) Calculate the coordinates of A, B and C. (iii) Calculate the area of triangle OBC where O is the origin. (iv) Find the coordinates of the point E such that ABEC is a parallelogram.
PS
PS
PS
PS
PS
PS
Exercise 5.
The intersection of two lines
4 The circle
You are, of course, familiar with the circle, and have done calculations involving its area and circumference. In this section you are introduced to the equation of a circle.
The circle is defi ned as the locus of all the points in a plane which are at a fi xed distance (the radius) from a given point (the centre).
This definition allows you to find the equation of a circle. Remember, the length of a line joining ( x 1 , y 1 ) to ( x 2 , y 2 ) is given by
length = x (^) 2 x (^) 1 ( y y )
2 2 1 ( − ) − −^2
For a circle of radius 3, with its centre at the origin, any point ( x , y ) on the circumference is distance 3 from the origin. So the distance of ( x , y ) from (0, 0) is given by
( x^ −^0 )^2 +^ ( y^ −0)^2 = 3
⇒ x^2 + y^2 = 3^2 ⇒ x^2 + y^2 = 9 This is the equation of the circle in Figure 5.22. The circle in Figure 5.23 has a centre (9, 5) and radius 4, so the distance between any point on the circumference and the centre (9, 5) is 4.
O x
y ( x – 9)^2 + ( y – 5)^2 = 4^2
( x – 9)
( y – 5)
(9, 5)
4
( x , y )
Figure 5.
The equation of the circle in Figure 5.23 is:
( x − 9 )^2 + ( y − 5)^2 = 4
⇒ ( x – 9) 2 + ( y – 5)^2 = 16.
prior knowledge
You should be able to complete the square, which is covered in Chapter 3. Locus means possible positions subject to given conditions. In two dimensions the locus can be a path or a region.
This is just Pythagoras’ theorem.
x
y
O
3 y x
( x , y )
x^2 + y^2 = 3^2
Figure 5.
Squaring both sides.
The circle
TeCHnology
Graphing software needs to be set to equal aspect to get these graphs looking correct.
5
Chapter 5 Coordinate geometry
ACTIVITY 5.
Sophie tries to draw the circle x^2 + y^2 = 9 on her graphical calculator. Explain what has gone wrong for each of these outputs.
(i)
1
2
4
6
–1 2
–3 –2 3
y
x
Figure 5.
T
Figure 5.
1
1
2
3
–1 2
–3 –2 3
y
x
(ii)
These results can be generalised to give the equation of a circle as follows: centre (0, 0), radius r : x^2 + y^2 = r^2 centre ( a , b ), radius r : ( x – a )^2 + ( y – b )^2 = r^2.
ACTIVITY 5.
Show that you can rearrange ( x − a )^2 + ( y − b )^2 = r^2 to give x^2 + y^2 − 2 ax − 2 by + ( a^2 + b^2 − r^2 ) = 0
note
In the form shown in the activity, the equation highlights some of the important characteristics of the equation of a circle. In particular: (i) the coeffi cients of x^2 and y^2 are equal (ii) there is no xy term.
Example 5.
Solution You need to rewrite the equation so it is in the form ( x – a )^2 + ( y – b )^2 = r^2. x^2 − 6 x + y^2 + 10 y − 15 = 0
( x − 3 ) 2 − 9 + ( y + 5) 2 − 25 − 15 = 0
( x − 3) 2 + ( y + 5) 2 = 49
So the centre is (3, −5) and the radius is 7.
Complete the square on the terms involving x …
Find the centre and radius of the circle (^) x^2 + y^2 − 6 x + 10 y − 15 = 0.
… then complete the square on the terms involving y.
5
Chapter 5 Coordinate geometry
Example 5.9 (^) (i) Show that OB is a diameter of the circle which passes through the
points O(0, 0), A(2, 6) and B(8, 4). (ii) Find the equation of the circle.
Solution (i) y
O^ x
C
B (8, 4)
A (2, 6)
Figure 5. If OB is the diameter of the circle, and A lies on the circle then (^) ∠OAB is 90°. So to show OB is the diameter you need to show that OA and AB are perpendicular.
Gradient of OA = 62 = 3
Gradient of AB = 62 −− 48 = (^) −^26 = – (^13)
Product of gradients = 3 × – 13 = – ⇒ Lines OA and AB are perpendicular so angle OAB = 90°. ⇒ OAB is the angle in a semicircle where OB is the diameter, as required.
(ii) The centre C of the circle is the midpoint of OB.
C = ( 0 2 +^8 , 0 2 +^4 ) = (4, 2)
The radius of the circle, CO = (^4 2) + 2 2 = 20. So the radius, r = 20 ⇒ r^2 = 20 Hence the equation of the circle is ( x − 4) 2 + ( y − 2)^2 = 20.
Always draw a sketch.
The angle in a semicircle is a right angle.
by the converse of the theorem that the angle in a semicircle is a right angle
To find the equation of a circle you need the centre and radius.
Let the centre be the point ( a , 4). Using Pythagoras’ theorem a^2 + 16 = 25 ⇒ a^2 = 9 ⇒ a = 3 or a = −3. The two possible equations are therefore ( x − 3)^2 + ( y − 4)^2 = 25 and ( x + 3)^2 + ( y − 4)^2 = 25.
The radius of the circle is 5 and the circle passes through the origin …
… so the distance between the centre ( a , 4) and the origin is 5.
( x − (−3))^2 + ( y − 4)^2 = 25
Example 5.10 (^) Figure 5.31 shows the circle
x^2 + y^2 = 25. The point P(4, 3) lies on the circle and the tangent to the circle at P cuts the coordinate axes at the points Q and R. Find (i) the equation of the tangent to the circle at P (ii) the exact area of triangle OQR.
Solution
(i) The gradient of OP is 34.
So the gradient of the tangent is (^) − 43.
The equation of the tangent at P(4, 3) is
y x y x x y
y x
y x x y
y x
y x x y
(ii) OQR forms a right-angled triangle. Find Q: − =
⇒ =
y
y
y
y
Find R: − =
⇒ =
x
x
x
x
Area of triangle OQR is 1 2
× × = 24 square units.
OP is the radius of the circle …
To find the equation of the tangent you need the gradient.
… the tangent and radius meet at right angles …
… so the gradient of the tangent is the negative reciprocal of the gradient of the radius.
Area is 12 × base × height.
The base is the x coordinate of R and the height is the y coordinate of Q.
Substitute x (^) = 0 into the tangent equation to find Q.
Substitute y (^) = 0 into the tangent equation to find R.
Exact means leave your answer as a fraction (or a surd).
x
y
O
P (4, 3)
Q
R
Figure 5.
The circle
