Chapter 1 Coordinates, points and lines

Transcription

1 Cambridge Universit Press Cambridge International AS and A Level Mathematics: Pure Mathematics Coursebook Hugh Neill, Douglas Quadling, Julian Gilbe Ecerpt Chapter Coordinates, points and lines This chapter uses coordinates to describe points and lines in two dimensions. When ou have completed it, ou should be able to find the distance between two points find the mid-point of a line segment, given the coordinates of its end points find the gradient of a line segment, given the coordinates of its end points find the equation of a line though a given point with a given gradient find the equation of the line joining two points recognise lines from different forms of their equations find the point of intersection of two lines tell from their gradients whether two lines are parallel or perpendicular. in this web service Cambridge Universit Press

2 Cambridge Universit Press Cambridge International AS and A Level Mathematics: Pure Mathematics Coursebook Hugh Neill, Douglas Quadling, Julian Gilbe Ecerpt Cambridge International AS and A Level Mathematics: Pure Mathematics. The distance between two points When ou choose an origin, draw an -ais to the right on the page and a -ais up the page and choose scales along the aes, ou are setting up a coordinate sstem. The coordinates of this sstem are called cartesian coordinates after the French mathematician René Descartes, who lived in the 7th centur. In Fig.., two points A and B have cartesian coordinates (4,3) and (0,7). The part of the line AB which lies between A and B is called a line segment. The length of the line segment is the distance between the points. ( 4,3) A 6 C B ( 0,7) 4 ( 0,3) A third point C has been added to Fig.. to form a right-angled triangle. You can see that C has the same -coordinate as B and the same -coordinate as A; that is, C has coordinates (0,3). Fig.. It is eas to see that AC has length 0 4 = 6, and CB has length 7 3 = 4. Using Pthagoras theorem in triangle ABC shows that the length of the line segment AB is ( 4) ( 7 ) You can use our calculator to give this as 7..., if ou need to, but often it is better to leave the answer as 5. The idea of coordinate geometr is to use algebra so that ou can do calculations like this when A and B are an points, and not just the particular points in Fig... It often helps to use a notation which shows at a glance which point a coordinate refers to. One wa of doing this is with sufies, calling the coordinates of the irst point (, ), and the coordinates of the second point (, ). Thus, for eample, stands for the -coordinate of the irst point. Fig.. shows this general triangle. You can see that C now has coordinates (, ), and that AC = and CB =. Pthagoras theorem now gives AB = ( ) + ( ). B (, ) An advantage of using algebra is that this formula works whatever the shape and position of the triangle. In Fig..3, the coordinates of A are negative, and in Fig..4 the line slopes downhill rather than uphill as ou move from left to right. Use Figs..3 and.4 to work out for ourself the length of AB in each case. You can then use the formula to check our answers. (, ) A C (, ) B3,5 ( ) A,5 ( ) C Fig.. B6, ( ) A(, ) C Fig..3 Fig..4 In Fig..3, so = 3 ( ) = 3 = 5 and 5 ( ) 5 + 6, AB = ( ( )) +( ( ( )) = 5 6 = = 6. in this web service Cambridge Universit Press

3 Cambridge Universit Press Cambridge International AS and A Level Mathematics: Pure Mathematics Coursebook Hugh Neill, Douglas Quadling, Julian Gilbe Ecerpt Chapter : Coordinates, points and lines And in Fig..4, so = 6 = 5 and 5 3, AB = ( ) ( +( 5) = 5 + ( 3) Also, it doesn t matter which wa round ou label the points A and B. If ou think of B as the irst point (, ) and A as the second point (, ), the formula doesn t change. For Fig.., it would give BA = ( ) +( ( 7) = ( 6) +( ( ) = = 5, as before. The distance between the points (, ) and (, ) (or the length of the line segment joining them) is ( ) + ( ).. The mid-point of a line segment You can also use coordinates to ind the mid-point of a line segment. Fig..5 shows the same line segment as in Fig.., but with the mid-point M added. The line through M parallel to the -ais meets AC at D. Then the lengths of the sides of the triangle ADM are half of those of triangle ACB, so that = AD = AC = ( 0 4 ) () 6 = 3, DM = CB = ( 7 3) = () 4 =. The -coordinate of M is the same as the -coordinate of D, which is 4 + AD = 4 ( 0 4) = 4 3= 7. The -coordinate of M is 3 3+ ( 7 3) = 3 = 5. So the mid-point M has coordinates (7,5). In Fig..6 points M and D have been added in the same wa to Fig... Eactl as before, AD AC ( ), DM CB ( ). So the -coordinate of M is DM + AD = + ( ) = + = + ( + ). The -coordinate of M is = = DM + ( ) = + = + ( + ). ( 4,3) A M B (, ) (, ) (, ) A D C M Fig..5 Fig..6 D C B ( 0,7) ( 0,3) 3 in this web service Cambridge Universit Press

4 Cambridge Universit Press Cambridge International AS and A Level Mathematics: Pure Mathematics Coursebook Hugh Neill, Douglas Quadling, Julian Gilbe Ecerpt Cambridge International AS and A Level Mathematics: Pure Mathematics The mid-point of the line segment joining (, ) and (, ) has coordinates ( ) + + ( + ), ( +. Now that ou have an algebraic form for the coordinates of the mid-point M ou can use it for an two points. For eample, for Fig..3 the mid-point of AB is ( (( ) 3 ), ( 5 ) ) ( ( 4 ) = ( ) ( ) ), (),. And for Fig..4 it is ( ( 6 )+ 3), ( 5 ) ) ( () 7, ( 7) ) (, 3 ) 3. Again, it doesn t matter which ou call the irst point and which the second. In Fig..5, if ou take (, ) as (0,7) and (, ) as (4,3), ou ind that the mid-point is ( 0 4), ( (, 5), as before. ( ) ). 3 The gradient of a line segment The gradient of a line is a measure of its steepness. The steeper the line, the larger the gradient. 4 Unlike the distance and the mid-point, the gradient is a propert of the whole line, not just of a particular line segment. If ou take an two points on the line and ind the increases in the - and -coordinates as ou go from one to the other, as in Fig..7, then the value of the fraction -step -step -step -step Fig..7 is the same whichever points ou choose. This is the gradient of the line. In Fig.. the -step and -step are and, so that: The gradient of the line joining (, ) to (, ) is. This formula applies whether the coordinates are positive or negative. In Fig..3, for eample, the gradient of AB is 5 ( ) = 3 =. ( ) 5 But notice that in Fig..4 the gradient is = = ; the negative gradient tells ou that the line slopes downhill as ou move from left to right. As with the other formulae, it doesn t matter which point has the sufi and which has the sufi. In Fig.., ou can calculate the gradient as either = 6 = 3,or 4 0 = 6 = 3. Two lines are parallel if the have the same gradient. in this web service Cambridge Universit Press

5 Cambridge Universit Press Cambridge International AS and A Level Mathematics: Pure Mathematics Coursebook Hugh Neill, Douglas Quadling, Julian Gilbe Ecerpt Chapter : Coordinates, points and lines EXAMPLE.3. The ends of a line segment are (p q, p + q) and (p + q, p q). Find the length of the line segment, its gradient and the coordinates of its mid-point. For the length and gradient ou have to calculate = ( p q ) ( p q ) = p q p q = q and = ( p q ) ( p q ) = p q p q = q. The length is ( ( q ) + ) = ( ) ( q) = 4q q 4q = 8 q. The gradient is q = =. q For the mid-point ou have to calculate + = ( p q ) ( p q) = p q + p q = p and + = ( p q ) + ( p q) = p q + p q = p. The mid-point is ( ( ), + ) ( ) ( (p ) ), ( p) = ( pp, ). Tr drawing our own igure to illustrate the results in this eample. EXAMPLE.3. 5 Prove that the points A(,), B(5,3), C(3,0) and D(, ) form a parallelogram. You can approach this problem in a number of was, but whichever method ou use, it is worth drawing a sketch. This is shown in Fig..8. Method (using distances) In this method, ind the lengths of the opposite sides. If both pairs of opposite sides are equal, then ABCD is a parallelogram. B5,3 ( ) AB = ( 5 ) +( ( ) = 0. DC = ( ( )) +( ( ( )) = 0. CB = ( 5 ) + ( 3 0) = 3. DA = ( ( )) + ( ( )) = 3. Therefore AB = DC and CB = DA, so ABCD is a parallelogram. D(, ) A, ( ) Fig..8 C3,0 ( ) Method (using mid-points) In this method, begin b inding the mid-points of the diagonals AC and BD. If these points are the same, then the diagonals bisect each other, so the quadrilateral is a parallelogram. The mid-point of AC is ( ( ), (+ 0) ), which is (, ). The mid-point of BD is ( ( 5 ( ), ( 3 +( ( )) ), which is also (, ). So ABCD is a parallelogram. in this web service Cambridge Universit Press

6 Cambridge Universit Press Cambridge International AS and A Level Mathematics: Pure Mathematics Coursebook Hugh Neill, Douglas Quadling, Julian Gilbe Ecerpt Cambridge International AS and A Level Mathematics: Pure Mathematics Method 3 (using gradients) In this method, ind the gradients of the opposite sides. If both pairs of opposite sides are parallel, then ABCD is a parallelogram. The gradients of AB and DC are 3 5 = 0 4 = ( ) and 3 ( ) = 4 = respectivel, so AB is parallel to DC. The gradients of DA and CB are both 3, so DA is parallel to CB. As the opposite sides are parallel, ABCD is a parallelogram. Eercise A Do not use a calculator. Where appropriate, leave square roots in our answers. Find the lengths of the line segments joining these pairs of points. In parts e and h assume that a > 0; in parts j and i assume that p > q > 0. a (, 5) and (7, 7) b ( 3, ) and (, ) c (4, 5) and (, 0) d ( 3, 3) and ( 7, 3) e (a, a) and (0a, 4a) f (a +, a + 3) and (a, a ) g (, 9) and (, 4) h (a, 5b) and (3a, 5b) i (p, q) and (q, p) j (p + 4q, p q) and (p 3q, p) Show that the points (, ), (6, ), (9, 3) and (4, ) are vertices of a parallelogram. 6 3 Show that the triangle formed b the points ( 3, ), (, 7) and (, 5) is isosceles. 4 Show that the points (7, ), ( 3, ) and (4, 5) lie on a circle with centre (, 0). 5 Find the coordinates of the mid-points of the line segments joining these pairs of points. a (, ), (6, 5) b (5, 7), ( 3, 9) c (, 3), (, 6) d ( 3, 4), ( 8, 5) e (p +, 3p ), (3p + 4, p 5) f (p + 3, q 7), (p + 5, 3 q) g (p + q, p + 3q), (5p q, p 7q) h (a + 3, b 5), (a + 3, b + 7) 6 A (, ) and B (6, 5) are the opposite ends of the diameter of a circle. Find the coordinates of its centre. 7 M(5, 7) is the mid-point of the line segment joining A (3, 4) to B. Find the coordinates of B. 8 A(, ), B(6, ), C(9, 3) and D(4, ) are the vertices of a parallelogram. Verif that the mid-points of the diagonals AC and BD coincide. 9 Which one of the points A(5, ), B(6, 3) and C(4, 7) is the mid-point of the other two? Check our answer b calculating two distances. 0 Find the gradients of the lines joining the following pairs of points. a (3, 8), (5, ) b (, 3), (, 6) c ( 4, 3), (0, ) d ( 5, 3), (3, 9) in this web service Cambridge Universit Press

7 Cambridge Universit Press Cambridge International AS and A Level Mathematics: Pure Mathematics Coursebook Hugh Neill, Douglas Quadling, Julian Gilbe Ecerpt Chapter : Coordinates, points and lines e (p + 3, p 3), (p + 4, p 5) f (p + 3, q 5), (q 5, p + 3) g (p + q, q + p 3), (p q +, q p + 3) h (7, p), (, p) Find the gradients of the lines AB and BC where A is (3, 4), B is (7, 6) and C is ( 3, ). What can ou deduce about the points A, B and C? The point P(, ) lies on the straight line joining A(3, 0) and B(5, 6). Find epressions for the gradients of AP and PB. Hence show that = A line joining a verte of a triangle to the mid-point of the opposite side is called a median. Find the length of the median AM in the triangle A(, ), B(0, 3), C(4, 7). 4 A triangle has vertices A(, ), B(3, 4) and C(5, 7). a Find the coordinates of M, the mid-point of AB, and N, the mid-point of AC. b Show that MN is parallel to BC. 5 The points A(, ), B(, 7) and C( 4, ) form a triangle. M is the mid-point of AB and N is the mid-point of AC. a Find the lengths of MN and BC. b Show that BC = MN. 6 The vertices of a quadrilateral ABCD are A(, ), B(7, 3), C(9, 7) and D( 3, 3). The points P, Q, R and S are the mid-points of AB, BC, CD and DA respectivel. a Find the gradient of each side of PQRS. b What tpe of quadrilateral is PQRS? 7 7 The origin O and the points P(4, ), Q(5, 5) and R(, 4) form a quadrilateral. a Show that OR is parallel to PQ. b Show that OP is parallel to RQ. c Show that OP = OR. d What shape is OPQR? 8 The origin O and the points L(, 3), M(4, 7) and N(6, 4) form a quadrilateral. a Show that ON = LM. b Show that ON is parallel to LM. c Show that OM = LN. d What shape is OLMN? 9 The vertices of a quadrilateral PQRS are P(, ), Q(7, 0), R(6, 4) and S( 3, ). a Find the gradient of each side of the quadrilateral. b What tpe of quadrilateral is PQRS? 0 The vertices of a quadrilateral are T(3, ), U(, 5), V(8, 7) and W(6, ). The midpoints of UV and VW are M and N respectivel. Show that the triangle TMN is isosceles. The vertices of a quadrilateral DEFG are D(3, ), E(0, 3), F(, 3) and G(4, ). a Find the length of each side of the quadrilateral. b What tpe of quadrilateral is DEFG? in this web service Cambridge Universit Press

8 Cambridge Universit Press Cambridge International AS and A Level Mathematics: Pure Mathematics Coursebook Hugh Neill, Douglas Quadling, Julian Gilbe Ecerpt Cambridge International AS and A Level Mathematics: Pure Mathematics The points A(, ), B(6, 0) and C(0, ) form an isosceles triangle with AB and BC of equal length. The point G is (6, 4). a Write down the coordinates of M, the mid-point of AC. b Show that BG = GM and that BGM is a straight line. c Write down the coordinates of N, the mid-point of BC. d Show that AGN is a straight line and that AG = GN..4 What is meant b the equation of a straight line or of a curve? How can ou tell whether or not the points (3, 7) and (, 5) lie on the curve = 3 +? The answer is to substitute the coordinates of the points into the equation and see whether the it; that is, whether the equation is satisied b the coordinates of the point. For (3, 7): the right side is = 9 and the left side is 7, so the equation is not satisied. The point (3, 7) does not lie on the curve = 3 +. For (, 5): the right side is 3 + = 5 and the left side is 5, so the equation is satisied. The point (, 5) lies on the curve = 3 +. The equation of a line or curve is a rule for determining whether or not the point with coordinates (,) lies on the line or curve. 8 This is an important wa of thinking about the equation of a line or curve..5 The equation of a line EXAMPLE.5. Find the equation of the line with gradient which passes through the point (, ). Fig..9 shows the line of gradient through A(, ), with another point P(, ) ling on it. P lies on the line if (and onl if) the gradient of AP is. The gradient of AP is. Equating this to gives =, which is = 4, or = 3. A(,) Fig..9 gradient P(,) In the general case, ou need to ind the equation of the line with gradient m through the point A with coordinates (, ). Fig..0 shows this line and another point P with coordinates (, ) on it. The gradient of AP is. Equating to m gives = m, or = m( ). The equation of the line through (, ) with gradient m is = m( ). Notice that the coordinates of A(, ) satisf this equation. A(, ) Fig..0 gradient m P(,) in this web service Cambridge Universit Press

9 Cambridge Universit Press Cambridge International AS and A Level Mathematics: Pure Mathematics Coursebook Hugh Neill, Douglas Quadling, Julian Gilbe Ecerpt Chapter : Coordinates, points and lines EXAMPLE.5. Find the equation of the line through the point (, 3) with gradient. Using the equation = m( ) gives the equation 3 = ( ( )), which is 3 = or = +. As a check, substitute the coordinates (, 3) into both sides of the equation, to make sure that the given point does actuall lie on the line. EXAMPLE.5.3 Find the equation of the line joining the points (3, 4) and (, ). To ind this equation, irst ind the gradient of the line joining (3,4) to (,). Then ou can use the equation = m( ). The gradient of the line joining (3, 4) to (, ) is 4 3 = 4 =. ( ) The equation of the line through (3,4) with gradient is ( 3). After multipling out and simplifing ou get 8 = 3, or = Check this equation mentall b substituting the coordinates of the other point..6 Recognising the equation of a line The answers to Eamples can all be written in the form = m + c, where m and c are numbers. It is eas to show that an equation of this form is the equation of a straight line. If = m + c, then c = m( 0), or 9 c 0 = m ( p = 0 ). This equation tells ou that, for all points (, ) whose coordinates satisf the equation, the line joining (0, c) to (, ) has gradient m. That is, (, ) lies on the line through (0, c) with gradient m. The point (0, c) lies on the -ais. The number c is called the -intercept of the line. c To ind the -intercept, put = 0 in the equation, which gives =. m But notice that ou can t do this division if m = 0. In that case the line is parallel to the -ais, so there is no -intercept. When m = 0, all the points on the line have coordinates of the form (something, c). Thus the points (, ), (, ), (5, ),... all lie on the straight line =, shown in Fig... As a special case, the -ais has equation = 0. Similarl, a straight line parallel to the -ais has an equation of the form = k. All points on it have coordinates (k, something). Thus the points (3, 0), (3, ), (3, 4),... all lie on the line = 3, shown in Fig... The -ais itself has equation = 0. (,) (, ) ( 5, ) Fig.. ( 3, 4) ( 3, ) ( 3, 0) The straight line = The straight line = 3 Fig.. in this web service Cambridge Universit Press

10 Cambridge Universit Press Cambridge International AS and A Level Mathematics: Pure Mathematics Coursebook Hugh Neill, Douglas Quadling, Julian Gilbe Ecerpt Cambridge International AS and A Level Mathematics: Pure Mathematics The line = k does not have a gradient; its gradient is undeined. Its equation cannot be written in the form = m + c..7 The eq uation a + b + c = 0 4 Suppose ou have the equation = +. It is natural to multipl b 3 to get = + 4, which can be rearranged to get = 0. This equation is in the form a + b + c = 0 where a, b and c are constants. Notice that the straight lines = m + c and a + b + c = 0 both contain the letter c, but it doesn t have the same meaning. For = m + c, c is the -intercept, but there is no similar meaning for the c in a + b + c = 0. A simple wa to ind the gradient of a + b + c = 0 is to rearrange it into the form =. Here are some eamples. EXAMPLE.7. 0 Find the gradient of the line = 0. Write this equation in the form =..., and then use the fact that the straight line = m + c has gradient m. From = 0 ou ind that 3 = + 4 and = 3 + comparing this equation with = m + c, the gradient is Therefore, EXAMPLE.7. One side of a parallelogram lies along the straight line with equation = 0. The point (,3) is a verte of the parallelogram. Find the equation of one other side. The line = 0 is the same as , so its gradient is 3 4. The line through (, 3) with gradient 3 4 is ( ), or = 0..8 The point of intersection of two lines Suppose that ou have two lines with equations = 4 and 3 + =. How do ou ind the coordinates of the point of intersection of these lines? You want the point (, ) which lies on both lines, so the coordinates (, ) satisf both equations. Therefore ou need to solve the equations simultaneousl. From these two equations, ou ind =, =, so the point of intersection is (, ). This argument applies to straight lines with an equations provided the are not parallel. To ind points of intersection, solve the equations simultaneousl. The method can also be used to ind the points of intersection of two curves. in this web service Cambridge Universit Press