A place for everthing, and everthing in its place. samuel smiles (8 904) Coordinate geometr Figure. shows some scaffolding in which some of the horizontal pieces are m long and others are m. All the vertical pieces are m. An ant crawls along the scaffolding from point P to point Q, travelling either horizontall or verticall. How far does the ant crawl? A mouse also goes from point P to point Q, travelling either horizontall or along one of the sloping pieces. How far does the mouse travel? A bee flies directl from point P to point Q. How far does the bee fl? Q P Figure. 6
Working with coordinates TeCHnolog When working through this chapter, ou ma wish to use a graphical calculator or graphing software to check our answers where appropriate. Working with coordinates Coordinates are a means of describing a position relative to a fied point, or origin. In two dimensions ou need two pieces of information; in three dimensions ou need three pieces of information. In the Cartesian sstem (named after René Descartes), position is given in perpendicular directions:, in two dimensions;,, z in three dimensions. This chapter concentrates eclusivel on two dimensions. The midpoint and length of a line segment When ou know the coordinates of two points ou can work out the midpoint and length of the line segment which connects them. ACTIVITY. Find (i) the coordinates of the midpoint, M (ii) the length AB. M B (8, ) Draw a right-angled triangle with AB as the hpotenuse and use Pthagoras theorem. A (, ) Figure. You can generalise these methods to find the midpoint and length of an line segment AB. Let A be the point (, ) and B the point (, ). (i) Find the midpoint of AB. The midpoint of two values is the mean of those values. The mean of the coordinates is +. The mean of the coordinates is +. So the coordinates of the midpoint are + +,. (ii) Find the length of AB. Figure.3 First find the lengths of AC and AB: AC = BC = B Pthagoras theorem: AB = AC + BC = ( ) + ( ) So the length AB is ( ) + ( ) A (, ) C has the same coordinate as B and the same coordinate as A. B (, ) C (, ) 66
Discussion point Does it matter which point ou call (, ) and which (, )? The gradient of a line When ou know the coordinates of an two points on a straight line, then ou can draw that line. The slope of a line is given b its gradient. The gradient is often denoted b the letter m. A (, 4) Figure.4 θ B (6, 7) C In Figure.4, A and B are two points on the line. The gradient of the line AB is given b the increase in the coordinate from A to B divided b the increase in the coordinate from A to B. In general, when A is the point (, ) and B is the point (, ), the gradient is m = 7 4 = 3 6 = 4 Gradient m = change in Gradient = change in 7 4 = 3 6 4 θ (theta) is the Greek letter th. α (alpha) and β (beta) are also used for angles. When the same scale is used on both aes, m = tanθ (see Figure.4). Chapter Coordinate geometr parallel and perpendicular lines ACTIVITY. It is best to use squared paper for this activit. Draw the line L joining (0, ) to (4, 4). Draw another line L perpendicular to L from (4, 4) to (6, 0). Find the gradients m and m of these two lines. What is the relationship between the gradients? Is this true for other pairs of perpendicular lines? When ou know the gradients m and m, of two lines, ou can tell at once if the are either parallel or perpendicular see Figure.. m Lines for which m m = will onl look perpendicular if the same scale has been used for both aes. m Figure. So for perpendicular lines: m = and likewise, m = m m. parallel lines: m = m m m perpendicular lines: m m = So m and m are each the negative reciprocal of each other. 67
Working with coordinates Eample. A and B are the points (, ) and (6, 3) respectivel (see Figure.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 A (, ) B (6, 3) Draw a diagram to help ou. Figure.6 (i) Gradient mab = = 3 6 = A A B B Gradient is difference in coordinates divided b difference in coordinates. It doesn t matter which point ou use first, as long as ou are consistent! (ii) Length AB = ( ) + ( ) = (6 ) + (3 ) = = 6 + 4 0 (iii) Midpoint = A + B A + B, = + 6, + ( 3 ) = (4,4) B A B A (iv) Gradient of AB: m AB = So gradient of perpendicular to AB is. Check: = The gradient of the line perpendicular to AB is the negative reciprocal of m AB. 68
Eample. The points P(, 7), Q(3, ) and R(0, ) form a triangle. (i) Use gradients to show that RP and RQ are perpendicular. (ii) Use Pthagoras theorem to show that PQR is right-angled. Solution R (0, ) P (, 7) Q (3, ) Alwas start b drawing a diagram. Chapter Coordinate geometr Figure.7 (i) Show that the gradients satisf m m = Gradient of RP = 0 = 7 Gradient of RQ = = 3 0 product of gradients = ( ) = sides RP and RQ are at right angles. (ii) Pthagoras theorem states that for a right-angled triangle with hpotenuse of length a and other sides of lengths b and c, a = b + c. Conversel, when a = b + c 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 hpotenuse. length = ( ) + ( ) PQ = (3 ) + ( 7) = + = 6 RP = ( 0) + (7 ) = 4 + 4 = 8 RQ = (3 0) + ( ) = 9 + 9 = 8 Since 6 = 8 + 8, PQ = RP + RQ PQ is the hpotenuse since RP and RQ are sides RP and RQ are at right angles. perpendicular. 69
Working with coordinates Eercise. For the following pairs of points A and B, calculate: (a) the midpoint of the line joining A to B (b) the distance AB (c) the gradient of the line AB (d) the gradient of the line perpendicular to AB. (i) A(, ) and B(6, 8) (ii) A(, ) and B( 6, 8) (iii) A(, ) and B(6, 8) (iv) A(, ) and B(6, 8) The gradient of the line joining the point P(3, 4) to Q(q, 0) is. Find the value of q. 3 The three points X(, ), Y(8, ) and Z(, ) are collinear. Find the value of. The lie on the same straight line. 4 For the points P(, ), and Q(3, ), find in terms of and : (i) the gradient of the line PQ (ii) the midpoint of the line PQ (iii) the length of the line PQ. The points A, B, C and D have coordinates (, ), (7, ), (9, 8) and (3, ) respectivel. (i) Find the gradients of the lines AB, BC, CD and DA. (ii) What do these gradients tell ou about the quadrilateral ABCD? (iii) Draw an accurate diagram to check our answer to part (ii). 6 The points A, B, and C have coordinates ( 4, ), (7, 4) and ( 3, ). (i) Draw the triangle ABC. (ii) Show b calculation that the triangle ABC is isosceles and name the two equal sides. (iii) Find the midpoint of the third side. (iv) Work out the area of the triangle ABC. 7 The points A, B and C have coordinates (, ), (b, 3) and (, ), where b > 3, and ABC = 90. Find: (i) the value of b (ii) the lengths of AB and BC (iii) the area of triangle ABC. 8 Three points A, B and C have coordinates (, 3), (3, ), and (, ).Find the value of in each of the following cases: (i) AB = AC (ii) AB = AC (iii) AB is perpendicular to BC (iv) A, B and C are collinear. 9 The triangle PQR has vertices P(8, 6), Q(0, ) and R(, r). Find the values of r when the triangle PQR: (i) has a right angle at P (ii) has a right angle at Q (iii) has a right angle at R (iv) is isosceles with RQ = RP. 0 A quadrilateral has vertices A(0, 0), B(0, 3), C(6, 6), and D(, 6). (i) Draw the quadrilateral. (ii) Show b calculation that it is a trapezium. (iii) EBCD is a parallelogram. Find the coordinates of E. Show that the points with coordinates (, ), (8, ), (7, 6) and (0, 0) are the vertices of a rhombus, and find its area. The lines AB and BC in Figure.8 are equal in length and perpendicular. gradient m A θ E D B gradient m Figure.8 (i) Show that triangles ABE and BCD are congruent. (ii) Hence prove that the gradients m and m satisf m m =. C 70
Each point on the line has an coordinate of 3. (c) Equations of the form = m = 4 These are lines through the origin, with gradient m. 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.9. (a) Equations of the form = a = = 3 (3, 0) (d) Equations of the form = m + c (0, ) All such lines are parallel to the ais. = (b) Equations of the form = b (0, ) = (0, ) These lines have gradient m and cross the ais at point (0, c). Each point on the line has a coordinate of. All such lines are parallel to the ais. (e) Equations of the form p + q + r = 0 This is often a tidier wa of writing the equation. + 3 6 = 0 Chapter Coordinate geometr Figure.9 (0, ) (, 0) (3, 0) = 3 + (3, 0) Eample.3 (i) Sketch the lines (a) = and (b) 3 + 4 = 4 on the same aes. (ii) Are these lines perpendicular? Solution (i) To draw a line ou need to find the coordinates of two points on it. (a) The line = passes through the point (0, ). Usuall it is easiest to find where the line cuts the and aes. The line is alread in the form = m + c. (b) Substituting = 0 gives =, so the line also passes through (, 0). Find two points on the line 3 + 4 = 4. Substituting = 0 gives 4 = 4 = 6 Set = 0 and find to give the -intercept. Then set = 0 and find to give the -intercept. substituting = 0 gives 3 = 4 = 8. So the line passes through (0, 6) and (8, 0). (Figure.0 overleaf) 7
The equation of a straight line (0, 6) = 6 4 3 3 + 4 = 4 (, 0) (8, 0) 0 3 4 6 7 8 (0, ) Warning When ou draw two perpendicular lines on a diagram, the will be at right angles if, and onl if, both aes are to the same scale. Figure.0 (ii) The lines look almost perpendicular but ou need to use the gradient of each line to check. Rearrange the equation Gradient of = is. to make the subject so Gradient of 3 + 4 = 4 is 3 4. ou can find the gradient. 4 = 3 + 4 Therefore the lines are not perpendicular as = 3 4 + 6 3 ( 4 ). Finding the equation of a line To find the equation of a line, ou need to think about what information ou are given. (i) Given the gradient, m, and the coordinates = m ( ) (, ) of one point on the line Take a general point (, ) on the line, as shown in Figure.. (, ) (, ) Figure. The gradient, m, of the line joining (, ) to (, ) is given b m = This is a ver useful form of the m ( = m) equation of a straight line. = For eample, the equation of the line with gradient that passes through the point (3, ) can be written as ( ) = ( 3) which can be simplified to = 7. 7
(ii) Given the gradient, m, and the -intercept (0, c) A special case of = m ( ) is when (, ) is the -intercept (0, c). = m + c The equation then becomes ACTIVITY.3 a Show algebraicall that an equivalent form of = = m + c Substituting = 0 and c = into the equation as shown in Figure.. When the line passes through the origin, the equation is = m The -intercept is (0, 0), so c = 0 as shown in Figure.3. (0, c) = m + c = m Chapter Coordinate geometr is =. B Use both forms to find the equation of the line joining (, 4) to (, 3) and show the give the same equation. Figure. Figure.3 (iii) Given two points, (, ) and (, ) The two points are used to find the gradient: m = = Discussion points How else can ou write the equation of the line? Which form do ou think is best for this line? This value of m is then substituted in the equation = m( ) This gives = ( ) or = (, ) Figure.4 (, ) (, ) 73
The equation of a straight line Eample.4 Find the equation of the line perpendicular to 4 + = which passes through the point P(, ). Solution First rearrange 4 + = into the form = m + c to find the gradient. 4 = + For perpendicular gradients m = m = + 3 So m = 4 m So the gradient is 4 The negative reciprocal of 4 is 4. Check: 4 = 4 So the gradient of a line perpendicular to = + 3 4 is 4. Using = m( ) when m = 4 and (, ) is (, ) ( ) = 4( ) + = 4 8 = 4 3 Straight lines can be used to model real-life situations. ften simplifing assumptions need to be made so that a linear model is appropriate. Eample. The diameter of a snooker cue varies uniforml from 9 mm to 3 mm over its length of 40 cm. Varing uniforml means that the graph of diameter against distance from the tip is a straight line. (i) Sketch the graph of diameter ( mm) against distance ( 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 mm. Solution (i) The graph passes through the points (0, 9) and (40, 3). diameter (mm) (40, 3) (0, 9) distance from tip (cm) Figure. 74
(ii) Gradient = = 3 9 40 0 = 0. Using the form = m + c, the equation of the line is = 0. + 9. (iii) Substituting = into the equation gives = 0. + 9 0. = 6 = 60 The diameter is mm at a point 60 cm from the tip. Discussion points Which of these situations in Figure.6 could be modelled b a straight line? For each straight line model, what information is given b the gradient of the line? What assumptions do ou need to make so that a linear model is appropriate? How reasonable are our assumptions? Chapter Coordinate geometr Interest earned on savings in a bank account against time Ta paid against earnings Mass of candle versus length of time it is burning Population of birds on an island against time Figure.6 Height of ball dropped from a cliff against time Cost of apples against mass of apples Distance travelled b a car against time Mobile phone bill against number of tets sent Profit of ice cream seller against number of sales Value of car against age of car Mass of gold bars against volume of gold bars Length of spring against mass of weights attached Eercise. Sketch the following lines: (i) (iii) = (ii) = = (iv) = + (v) = + (vii) = (vi) = (viii) + + = 0 Find the equations of the lines (i) (v) in Figure.7. 6 (iii) (ii) 4 (i) 4 0 4 6 8 4 (v) (iv) Figure.7 3 Find the equations of the lines (i) parallel to = 3 and passing through (0, 0) (ii) parallel to = 3 and passing through (, ) (iii) parallel to + 3 = 0 and passing through (, ) (iv) parallel to 3 = 0 and passing through (, ) (v) parallel to + = 3 and passing through (, ). 4 Find the equations of the lines (i) perpendicular to = 3 and passing through (0, 0) (ii) perpendicular to = + 3 and passing through (4, 3) (iii) perpendicular to + = 4 and passing through (4, 3) 7
The equation of a straight line (iv) perpendicular to = + and passing through ( 4, 3) (v) perpendicular to + 3 = 4 and passing through ( 4, 3). Find the equations of the line AB in each of the following cases. (i) A(3, ), B(, 7) (ii) A( 3, ), B(, 7) (iii) A( 3, ), B(, 7) (iv) A(3, ), B(, 7) (v) A(, 3), B(7, ) 6 Show that the region enclosed b the lines = + 3, = 3, + 3 + = 0 and + 3 + = 0 forms + a rectangle. + = The perpendicular bisector is the line at right angles to AB (perpendicular) that passes though the midpoint of AB (bisects). 7 Find the equation of the perpendicular bisector of each of the following pairs of points. (i) A(, 4) and B(3, ) (ii) (A(4, ) and B (, 3) (iii) A(, 4) and B( 3, ) (iv) A(, 4) and B( 3, ) (v) A(, 4) and B(3, ) 8 A median of a triangle is a line joining one of the vertices to the midpoint of the opposite side. In a triangle AB, is at the origin, A is the point (0, 6), and B is the point (6, 0). (i) Sketch the triangle. (ii) Find the equations of the three medians of the triangle. (iii) Show that the point (, ) lies on all three medians. (This shows that the medians of this triangle are concurrent.) 9 A quadrilateral ABCD has its vertices at the points (0, 0), (, ), (0, 0) and ( 6, 8) respectivel. (i) Sketch the quadrilateral. (ii) Find the gradient of each side. (iii) Find the length of each side. (iv) Find the equation of each side. (v) Find the area of the quadrilateral. 0 Afi rm manufacturing jackets finds that it is capable of producing 00 jackets per da, but it can onl sell all of these if the charge to the wholesalers is no more than 0 per jacket. n the other hand, at the current price of per jacket, onl 0 can be sold per da. Assuming that the graph of price P against number sold per da N is a straight line: (i) sketch the graph, putting the number sold per da on the horizontal ais (as is normal practice for economists) (ii) find its equation. Use the equation to find: (iii) the price at which 88 jackets per da could be sold (iv) the number of jackets that should be manufactured if the were to be sold at 3.70 each. To clean the upstairs window on the side of a house, it is necessar to position the ladder so that it just touches the edge of the lean-to shed as shown in Figure.8. The coordinates represent distances from in metres, in the and directions shown. Figure.8 (i) (ii) (iii) shed A ladder (., ) B (., 0) Find the equation of the line of the ladder. Find the height of the point A reached b the top of the ladder. Find the length of the ladder to the nearest centimetre. 76
A spring has an unstretched length of 0 cm. When it is hung with a load of 80 g attached, the stretched length is 8 cm. Assuming that the etension of the spring is proportional to the load: (i) draw a graph of etension E against load L and find its equation (ii) find the etension caused b a load of 48 g (iii) find the load required to etend the spring to a length of 0 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.) 3 (iv) Find the load which would cause this to happen. (0, b) Figure.9 (a, 0) Show that the equation of the line in Figure.9 can be written + =. a b 3 The intersection of two lines The intersection of an two curves (or lines) can be found b solving their equations simultaneousl. In the case of two distinct lines, there are two possibilities: (i) the are parallel, or (ii) the intersect at a single point. You often need to find where a pair of lines intersect in order to solve problems. Chapter Coordinate geometr Eample.6 Discussion point The line l has equation = 4 and the line m has equation = 3. What can ou sa about the intersection of these two lines? The lines = 3 and + 3 = 0 intersect at the point P. Find the coordinates of P. Solution You need to solve the equations = 3 and + 3 = 0 simultaneousl. Substitute equation into : ( 3) + 3 = 0 0 6 + 3 = 0 Multipl out the brackets. 3 6 = 0 Simplif 3 = 6 = Don t forget to find Substitute = into equation to find. the coordinate. = 3 = 3 So the coordinates of P are (, 3). 77
The intersection of two lines Eercise.3 Find the coordinates of the point of intersection of the following pairs of lines. (i) = + 3 and = 6 + (ii) = 3 and + = 4 (iii) 3 + = 4 and 4 = 3 (i) Find the coordinates of the points where the following pairs of lines intersect. (a) = 4 and = 7 (b) = + and = 7 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 b the lines AB: = BC: 7 + 6 = 3 and AC: 9 + =. (ii) Show that the triangle is isosceles. 4 A(0, ), B(, 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 tpe of quadrilateral is ABCD? The line = crosses the ais at A. The line = + 4 crosses the ais 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 eact area of the triangle ABP. 6 Triangle ABC has an angle of 90 at B. Point A is on the ais, AB is part of the line + 8 = 0, and C is the point (6, ). (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 our answer to (iii), find the length of the perpendicular from B to AC. 7 Two rival tai firms have the following fare structures: Firm A: fied charge of plus 40p per kilometre; Firm B: 60p per kilometre, no fied charge. (i) Sketch the graph of price (vertical ais) against distance travelled (horizontal ais) for each firm (on the same aes). (ii) Find the equation of each line. (iii) Find the distance for which both firms charge the same amount. (iv) Which firm would ou use for a distance of 6 km? 8 Two sides of a parallelogram are formed b parts of the lines = 9 and = 9. (i) Show these two lines on a graph. (ii) Find the coordinates of the verte where the intersect. Another verte of the parallelogram is the point (, ). (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 + = 0 meets the ais at A and the line with equation + = meets the ais 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 BC where is the origin. (iv) Find the coordinates of the point E such that ABEC is a parallelogram. 78
0 Figure.0 shows the suppl and demand of labour for a particular industr in relation to the wage paid per hour. Suppl is the number of people willing to work for a particular wage, and this increases as the wage paid increases. Demand is the number of workers that emploers are prepared to emplo at a particular wage: this is greatest for low wages. wage rate ( per hour) W 6 4 3 (000, ) (000, 3) (L, W) suppl demand 000 00 000 00 quantit of labour (person hours per week) (00, 6) (00, 3) L A median of a triangle is a line joining a verte to the midpoint of the opposite side. In an triangle, the three medians meet at a point called the centroid of the triangle. Find the coordinates of the centroid for each triangle shown in Figure.. (i) (0, ) Chapter Coordinate geometr Figure.0 (6, 0) (i) Find the equation of each of the lines. (ii) Find the values of L and W at which the market clears, i.e. at which suppl equals demand. (iii) Although economists draw the graph this wa round, mathematicians would plot wage rate on the horizontal ais. Wh? When the market price p of an article sold in a free market varies, so does the number demanded, D, and the number supplied, S. In one case D = 0 + 0.p and S = + p. (i) Sketch both of these lines on the same graph. (Put p on the horizontal ais.) The market reaches a state of equilibrium when the number demanded equals the number supplied. (ii) Find the equilibrium price and the number bought and sold in equilibrium. (ii) (0, 9) (, 0) Figure. (, 0) 3 Find the eact area of the triangle whose sides have the equations + = 4, = 8 and + =. 79
The circle prior knowledge You should be able to complete the square, which is covered in Chapter 3. 4 The circle You are, of course, familiar with the circle, and have done calculations involving its area and circumference. In this section ou are introduced to the equation of a circle. The circle is defined as the locus of all the points in a plane which are at a fied distance (the radius) from a given point (the centre). This definition allows ou to find the equation of a circle. Remember, the length of a line joining (, ) to (, ) is given b ( ) length = ( ) Locus means possible positions subject to given conditions. In two dimensions the locus can be a path or a region. This is just Pthagoras theorem. For a circle of radius 3, with its centre at the origin, an point (, ) on the circumference is distance 3 from the origin. So the distance of (, ) from (0, 0) is given b ( 0 ) + ( 0) 3 (, ) + = 3 + = 9 Squaring both sides. This is the equation of the circle in Figure.. Figure. The circle in Figure.3 has a centre (9, ) and radius 4, so the distance between an point on the circumference and the centre (9, ) is 4. + = 3 ( 9) + ( ) = 4 (, ) TeCHnolog Graphing software needs to be set to equal aspect to get these graphs looking correct. (9, ) 4 ( 9) ( ) Figure.3 The equation of the circle in Figure.3 is: ( 9 ) + ( ) = 4 ( 9) + ( ) = 6. 80
T ACTIVITY.4 Sophie tries to draw the circle + = 9 on her graphical calculator. Eplain what has gone wrong for each of these outputs. (i) 3 6 4 3 4 (ii) 3 3 3 Chapter Coordinate geometr 6 3 Figure.4 Figure. note In the form shown in the activit, the equation highlights some of the important characteristics of the equation of a circle. In particular: (i) the coefficients of and are equal (ii) there is no term. ACTIVITY. Show that ou can rearrange ( a) + ( b) = r to give + a b + (a + b r ) = 0 These results can be generalised to give the equation of a circle as follows: centre (0, 0), radius r : + = r centre (a, b), radius r : ( a) + ( b) = r. Eample.7 Find the centre and radius of the circle + 6 + 0 = 0. Solution You need to rewrite the equation so it is in the form ( a) + ( b) = r. 6 + + 0 = 0 Complete the square on the terms involving ( 3) 9 + ( + ) = 0 then complete the square on the terms ( 3) + ( + ) = 49 involving. So the centre is (3, ) and the radius is 7. 7 = 49 8
The circle Circle geometr There are some properties of a circle that are useful when solving coordinate geometr problems. The angle in a semicircle is a right angle (see Figure.6). Discussion points How can ou prove these results? State the converse of each of these results. The converse of p implies q is q implies p. The perpendicular from the centre of a circle to a chord bisects the chord (see Figure.7). 3 The tangent to a circle at a point is perpendicular to the radius through that point (see Figure.8). Figure.6 Figure.7 Figure.8 The converse of each of the three circle properties above is also true. The net three eamples use these results in coordinate geometr. Eample.8 A circle has a radius of units, and passes through the points (0, 0) and (0, 8). Sketch the two possible positions of the circle and find their equations. Solution (0, 8) The line joining (0, 0) to (0, 8) is a chord of each circle. The perpendicular bisector of a chord passes through the centre of each circle. Figure.9 The midpoint of the chord is (0, 4). The equation of the bisector is = 4. So the centre of the circle lies on the line = 4. The chord is along the ais, so the perpendicular bisector passes through (0, 4) and is parallel to the ais. 8
Eample.9 Let the centre be the point (a, 4). Using Pthagoras theorem a + 6 = a = 9 a = 3 or a = 3. The two possible equations are therefore (i) Show that B is a diameter of the circle which passes through the points (0, 0), A(, 6) and B(8, 4). (ii) Find the equation of the circle. Solution (i) ( 3) + ( 4) = and ( + 3) + ( 4) =. A (, 6) The radius of the circle is and the circle passes through the origin so the distance between the centre (a, 4) and the origin is. ( ( 3)) + ( 4) = Chapter Coordinate geometr B (8, 4) C Alwas draw a sketch. Figure.30 If B is the diameter of the circle, and A lies on the circle then AB is 90. The angle in a semicircle is a right angle. So to show B is the diameter ou need to show that A and AB are perpendicular. Gradient of A = 6 = 3 Gradient of AB = 6 4 8 = 6 = 3 Product of gradients = 3 3 = Lines A and AB are perpendicular so angle AB = 90. AB is the angle in a semicircle where B is the diameter, as required. (ii) The centre C of the circle is the midpoint of B. C = ( 0 + 8 0 + 4, ) = (4, ) To find the equation of a circle ou need the centre and radius. The radius of the circle, C = 4 + = 0. So the radius, r = 0 r = 0 b the converse of the theorem that the angle in a semicircle is a right angle Hence the equation of the circle is ( 4) + ( ) = 0. 83
The circle Eample.0 Figure.3 shows the circle + =. Q The point P(4, 3) lies on the circle and the tangent to the circle at P cuts the coordinate aes at the points Q and R. P (4, 3) Find (i) the equation of the tangent to the circle at P R (ii) the eact area of triangle QR. Figure.3 Solution (i) The gradient of P is 3 4. P is the radius of the circle To find the equation of the tangent ou need the gradient. So the gradient of the tangent is the tangent and radius meet at right angles 4. 3 The equation of the tangent at P(4, 3) is 3 4 3 = 3 ( 4) 3 9 = 6= 4 4 + 3 + = 0 = (ii) QR forms a right-angled triangle. Find Q: 3 = 0 = 3 Find R: 4 = 0 = 4 Area of triangle QR is 6 4 3 = 4 square units. Substitute = 0 into the tangent equation to find Q. Substitute = 0 into the tangent equation to find R. so the gradient of the tangent is the negative reciprocal of the gradient of the radius. Area is base height. The base is the coordinate of R and the height is the coordinate of Q. Eact means leave our answer as a fraction (or a surd). 84
Eercise.4 Find the equations of the following circles. (i) centre (, 3), radius (ii) centre (, 3), radius (iii) centre (, 3), radius 3 (iv) centre (, 3), radius 4 For each of the following circles state (a) the coordinates of the centre (b) the radius. (i) + = (ii) + ( ) = (iii) ( ) + = 3 (iv) ( + ) + ( + ) = 4 (v) ( ) + ( + ) = 3 The equation of a circle is ( 3) + ( + ) = 6. Complete the table to show whether each point lies inside the circle, outside the circle or on the circle. Point Inside utside n (3, ) (, ) (6, 6) (4, 3) (0, ) (, 3) 4 Draw the circles ( 4) + ( ) = 6 and ( 3) + ( 3) = 4. In how man points do the intersect? Sketch the circle ( + ) + ( 3) = 6, and find the equations of the four tangents to the circle which are parallel to the coordinate aes. 6 Find the coordinates of the points where each of these circles crosses the aes. (i) + = (ii) ( 4) + ( + ) = (iii) ( + 6) + ( 8) = 00 7 Find the equation of the circle with centre (, 7) passing through the point ( 4, ). 8 Show that the equation + + 4 + = 0 can be written in the form ( + ) + ( ) = r, where the value of r is to be found. Hence give the coordinates of the centre of the circle, and its radius. 9 Draw the circle of radius 4 units which touches the positive and aes, and find its equation. 0 A(3, ) and B(9, 3) lie on a circle. Show that the centre of the circle lies on the line with equation 4 3 + 4 = 0. For each of the following circles find (a) the coordinates of the centre (b) the radius. (i) + 6 6 = 0 (ii) + + + 6 6 = 0 (iii) + + 8 + 8 = 0 A circle passes through the points A(3, ), B(, 6) and C(, 3). (i) Calculate the lengths of the sides of the triangle ABC. (ii) Hence show that AC is a diameter of this circle. State which theorems ou have used, and in each case whether ou have used the theorem or its converse. (iii) Calculate the area of triangle ABC. 3 (i) Find the midpoint, C, of AB where A and B are (, 8) and (3, 4) respectivel. Find also the distance AC. (ii) Hence find the equation of the circle which has AB as a diameter. 4 A(, ) is a point on the circle ( 3) + ( + ) =. (i) State the coordinates of the centre of the circle and hence find the coordinates of the point B, where AB is a diameter of the circle. (ii) C(, ) also lies on the circle. Use coordinate geometr to verif that angle ACB = 90. The tangent to the circle + ( + 4) = at the point ( 4, ) intersects the ais at A and the ais at B. Find the eact area of the triangle AB. 6 A circle passes through the points (, 0) and (8, 0) and has the ais as a tangent. Find the two possible equations for the circle. 7 A(6, 3) and B(0,) are two points on a circle with centre (, 8). (i) Calculate the distance of the chord AB from the centre of the circle. (ii) Find the equation of the circle. 8 A(6, 6), B(6, ) and C(, ) are three points on a circle. Find the equation of the circle. Chapter Coordinate geometr 8
The intersection of a line and a curve prior knowledge You need to be able to solve a quadratic equation use the discriminant to determine the number of roots of a quadratic equation. These are covered in Chapter 3. The intersection of a line and a curve When a line and a curve are in the same plane, the coordinates of the point(s) of intersection can be found b solving the two equations simultaneousl. There are three possible situations. (i) All points of intersection are distinct (see Figure.3). = + = + 4 = 4 ( 4) + ( 3) = There are (or more) distinct solutions to the simultaneous equations. Figure.3 (ii) The line is a tangent to the curve at one (or more) point(s) (see Figure.33). In this case, each point of contact corresponds to two (or more) coincident points of intersection. It is possible that the tangent will also intersect the curve somewhere else (as in Figure.33b). (a) (b) = 3 + 6 ( 4) + ( 4) = 3 = + = (, 8) 3 When ou solve the simultaneous equations ou will obtain an equation with a repeated root. = Figure.33 (iii) The line and the curve do not meet (see Figure.34). When ou tr to solve the simultaneous equations ou will obtain an equation with no roots. So there is no point of intersection. = Figure.34 86
Eample. A circle has equation + = 8. note This eample shows ou an important result. When ou are finding the intersection points of a line and a quadratic curve, or two quadratic curves, ou obtain a quadratic equation. If the discriminant is: positive there are two points of intersection zero there is one repeated point of intersection negative there are no points of intersection. For each of the following lines, find the coordinates of an points where the line intersects the circle. (i) = (ii) = + 4 (iii) = + 6 Solution (i) Substituting = into + = 8 gives + = 8 = 8 Simplif. = 4 Don t forget the negative = ± square root! The line Since = then the coordinates are (, ) and (, ). intersects the circle twice. (ii) Substituting = + 4 into + = 8 gives + ( + 4) = 8 Multipl out the brackets. + + 8 + 6 = 8 + 8 + 8 = 0 + 4 + 4 = 0 Divide b. ( + ) = 0 This is a repeated root, so = = + 4 is a tangent to the circle. When = then = + 4 = So the coordinates are (, ). (iii) Substituting = + 6 into + = 8 gives + ( + 6) = 8 + + + 36 = 8 + + 8 = 0 Check the discriminant: b 4ac 4 8 = 80 Since the discriminant is less than 0, the equation has no real roots. So the line = + 6 does not meet the circle. Chapter Coordinate geometr The intersection of two curves The same principles appl to finding the intersection of two curves, but it is onl in simple cases that it is possible to solve the equations simultaneousl using algebra (rather than a numerical or graphical method). 87
The intersection of a line and a curve Eample. Sketch the circle + = 6 and the curve = 4 on the same aes. Find the coordinates of an points of intersection. Discussion points How else could ou solve the simultaneous equations in the eample? Which method is more efficient in this case? Solution The circle has centre (0, 0) and radius 4. 4 4 = 4 + = 6 The curve = 4 is the same shape as = but translated verticall down the ais b 4 units. B smmetr, the circle cuts the aes at (4, 0), ( 4, 0) and (0, 4). (0, 4) Figure.3 Substituting = 4 into + = 6 gives + ( 4) = 6 + 4 8 + 6 = 6 Simplif. 4 7 = 0 ( 7) = 0 Factorise. = 0 = 0 (twice). or = 7 = ± 7 Don t forget the negative square root. Substitute back into = 4 to find the coordinates. = 0 = 4 = ± 7 = 7 4 = 3 So the points of intersection are ( 7, 3), ( 7, 3) and (0, 4) (twice). Eercise. Show that the line = 3 + crosses the curve = + 3 at (, 4) and find the coordinates of the other point of intersection. Find the coordinates of the points where the line = cuts the circle ( ) + ( + ) =. 3 Find the coordinates of the points of intersection of the line = and the circle ( + ) + ( ) = 0. What can ou sa about this line and the circle? 4 (i) (i) Show that the line + = 6 is a tangent to the circle + = 8. (ii) Show the line and circle on a diagram. Find the point of contact of the tangent parallel to the line + = 6, and the equation of the tangent. Find the coordinates of the points of intersection of the line = and the curve = + 6. (ii) Show also that the line = does not cross the curve = + 6 +. 88
6 Find the coordinates of the points A and B where the line 3 + = 0 cuts the circle + + 6 + = 0. Also find the coordinates of the points where the line = + meets the curve = 3 3 + 3 +. 7 Figure.36 shows the cross-section of a goldfish bowl. The bowl can be thought of as a sphere with its top removed and its base flattened. 6 C 0 all dimensions in centimetres Figure.36 Assume the base is on the ais and the ais is a line of smmetr. (i) Find the height of the bowl. (ii) Find the equation of the circular part of the cross-section. (iii) The bowl is filled with water to a depth of cm. Find the area of the surface of the water. 8 The line = intersects the circle + = at two points A and B. (i) Find the coordinates of the points and the distance AB. (ii) Is AB a diameter of the circle? Give a reason for our answer. 9 (i) Find the value of k for which the line = + k forms a tangent to the curve =. (ii) Hence find the coordinates of the point where the line = + k meets the curve for the value of k found in part (i). 0 The equation of a circle is ( + ) + = 8 and the equation of a line is + = k, where k is a constant. Find the values of k for which the line forms a tangent to the curve. The equations of two circles are given below. ( + ) + ( ) = 0 and ( ) + ( 3) = The two circles intersect at the points A and B. Find the area of the triangle AB where is the origin. Chapter Coordinate geometr 89
Problem solving integer point circles 0 8 6 4 0 8 6 4 0 4 6 8 0 4 6 8 Figure.37 0 Look at the circle in Figure.37. Its equation is + = 00. It goes through the point (6, 8). Since both 6 and 8 are integers, this is referred to as an integer point in this question. This is not the onl integer point this circle goes through; another is ( 0, 0) and there are others as well. (i) How man integer points are there on the circle? (ii) How man circles are there with equations of the form + = N, where 0 < N < 00, that pass through at least one integer point? How man of these circles pass through at least integer points? (iii) Devise and eplain at least one method to find the equation of a circle with radius greater than 0 units that passes through at least integer points. 90
Problem specification and analsis Parts (i) and (ii) of the problem are well defined and so deal with them first. Start b thinking about possible strategies. There are several quite different approaches, based on geometr or algebra. You ma decide to tr more than one and see how ou get on. Part (iii) is more open ended. You have to devise and eplain at least one method. Leave this until ou get to the last stage of the problem solving ccle. B then our earlier work ma well have given ou some insight into how to go about it. Information collection In this problem there will probabl be a large amount of trial and error in our data collection. As well as collecting information, ou will be tring out different possible approaches. There are a number of cases that ou could tr out and so ou need to be on the lookout for patterns that will cut down on our work. You have to think carefull about how ou are going to record our findings sstematicall. 3 Processing and representation The work ou need to do at this stage will depend on what ou have alread done at the Information collection stage. You ma have alread collected all the information ou need to answer parts (i) and (ii) b just counting up the numbers. Alternativel, however, ou ma have found some patterns that will help ou to work out the answers. You then need to find a good wa to present our answers. Think of someone who is unfamiliar with the problem. How are ou going to show such a person what ou have found in a convincing wa? 4 Interpretation So far ou have been looking at parts (i) and (ii) of the problem. The are well defined and all the answers are numbers. In part (iii), ou are now epected to interpret what ou have been doing b finding not just numbers but also a method, so that ou can continue the work with larger circles. To give a good answer ou will almost certainl need to use algebra but ou will also need to eplain what ou are doing in words. The wording of the questions suggests there is more than one method and that is indeed the case. So a reall good answer will eplore the different possibilities. Chapter Coordinate geometr 9
The intersection of a line and a curve learning outcomes Now ou have finished this chapter, ou should be able to solve problems involving finding the midpoint of two points solve problems involving finding the distance between two points understand gradient as the rate of change find the gradient of a line joining two points recall and use the relationships between gradients for parallel and perpendicular lines solve problems with parallel and perpendicular lines draw or sketch a line, given its equation find the equation of a line solve real-life problems that can be modelled b a linear function find the intersection of two lines find the centre and radius of a circle from its equation when the equation of the circle is given in its standard form when the equation of the circle needs to be rewritten in completed square form find the equation of a circle given the radius and the centre using circle theorems to find the centre and radius find the equation of a tangent to a circle using the circle theorems find the points of intersection of a line and a curve understanding the significance of a repeated root in the case of a line which is a tangent to the curve understanding the significance of having no roots in the case of a line which does not intersect the curve find the intersection points of curves in simple cases. 9
Ke points For a line segment A(, ) and B(, ) (Figure.38) then: A (, ) Figure.38 n the gradient of AB is n the midpoint is + +, n the distance AB is ( ) + ( ). B (, ) Two lines are parallel their gradients are equal. 3 Two lines are perpendicular the product of their gradients is. 4 The equation of a straight line ma take an of the following forms: n line parallel to the ais: = a n line parallel to the ais: = b n line through the origin with gradient m: = m n line through (0, c) with gradient m: = m + c n line through (, ) with gradient m: = m( ) The equation of a circle is n centre (0, 0), radius r: + = r n centre (a, b), radius r: ( a) + ( b) = r. 6 The angle in a semicircle is a right angle (Figure.39). 7 The perpendicular from the centre of a circle to a chord bisects the chord (Figure.40). 8 The tangent to a circle at a point is perpendicular to the radius through that point (Figure.4). using Pthagoras theorem Chapter Coordinate geometr Figure.39 Figure.40 Figure.4 9 To find the points of intersection of two curves, ou solve their equations simultaneousl. 93
Practice questions Pure Mathematics PRACTICE QUESTINS PURE MAThEMATICS MP (i) Prove that = 3. [ marks] 3 MP MP (ii) Show that 3 + 3 = 3 +. [ marks] (i) Solve the equation 3 = 4 + 4. [3 marks] (ii) Find a value of which is a counter-eample to 0 >. [ mark] 3 Do not use a calculator in this question. Figure shows the curves = 4 + and = 7. = 7 = 4 + Figure (i) Find the coordinates of their points of intersection. [ marks] (ii) Prove that = is a tangent to = 4 + and state the coordinates of the point of contact. 4 Do not use a calculator in this question. [4 marks] (i) Write + 6 + 7 in the form ( + a) + b. [3 marks] (ii) (iii) State the coordinates of the turning point of = + 6 + 7 and whether it is a minimum or maimum. Sketch the curve = + 6 + 7 and solve the inequalit + 6 + 7 > 0. [3 marks] [4 marks] 94
Figure shows a circle with centre C which passes through the points A(, 4) and B(, ). A (, 4) B (, ) Figure C (i) AB is a chord of the circle. Show that the centre of the circle must lie on the line + = 3, eplaining our reasoning. (ii) The centre of the circle also lies on the ais. Find the equation of the circle. 6 Figure 3 shows an equilateral triangle, ABC with A and B on the ais and C on the ais. [7 marks] [ marks] Practice questions Pure Mathematics C G F A D E B Figure 3 Each side of triangle ABC measures 4 units. (i) Find the coordinates of points A, B and C in eact form. [4 marks] (ii) Show that the equation of line BC can be written as = 3( ). [ marks] A rectangle DEFG is drawn inside the triangle, as also shown in Figure 3. D and E lie on the ais, G on AC and F on BC. (iii) Find the greatest possible area of rectangle DEFG. [7 marks] 9
Practice questions Pure Mathematics M T 7 Figure 4 shows a spreadsheet with the information about tpical stopping distances for cars from the Highwa Code. Figure has been drawn using the spreadsheet. 0 3 4 6 7 Home A Speed (mph) Figure 4 0 30 40 0 60 70 Insert B Thinking distance (m) 6 9 8 Page Laout f C Braking distance (m) 6 4 4 38 7 Formulas D Total stopping distance (m) 3 36 3 73 96 Distance (m) 00 80 60 40 0 0 0 0 40 60 80 Speed (mph) Ke: thinking distance (m) braking distance (m) total stopping distance (m) Figure (i) (ii) (iii) (a) What feature of the scatter diagram (Figure ) suggests that the thinking distance is directl proportional to speed? [ mark] (b) What does this tell ou about the thinking time for different speeds? Comment, with a brief eplanation, on whether this is a reasonable modelling assumption. [ marks] (c) Write down a formula connecting the speed, mph and the thinking distance d m. [ mark] The spreadsheet (Figure 4) gives the following linear best fit model for the total stopping distance, m in terms of the speed mph. =.677 6.38 (a) Use the model to find the total stopping distance for a speed of 0 mph. [ mark] (b) Eplain wh this is not a suitable model for total stopping distance. [ mark] The spreadsheet gives the following quadratic best fit model for the total stopping distance. = 0.07 + 0.69 + 0.6 The values for total stopping distance using this model are shown in Table. Table Speed (mph) 0 30 40 0 60 70 Quadratic model (m).67 36.36.99 7.894 9.933 (a) Calculate the missing value for 0 mph. [ mark] (b) Give one possible reason wh the model does not give eactl the same total stopping distances as those listed in the Highwa Code. [ mark] 96