C1: Coordinate geometry of straight lines

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1 B_Chap0_08-05.qd 5/6/04 0:4 am Page 8 CHAPTER C: Coordinate geometr of straight lines Learning objectives After studing this chapter, ou should be able to: use the language of coordinate geometr find the distance between two given points find the coordinates of the mid-point of a line segment joining two given points find, use and interpret the gradient of a line segment know the relationship between the gradients for parallel and for perpendicular lines find the equations of straight lines given (a) the gradient and -intercept, (b) the gradient and a point, and (c) two points verif, given their coordinates, that points lie on a line find the coordinates of a point of intersection of two lines find the fourth verte of a parallelogram given the other three. Coordinate geometr is the use of algebraic methods to stud the geometr of straight lines and curves. In this chapter ou will consider onl straight lines. Curves will be studied later.. Cartesian coordinates In our GCSE course ou plotted points in two dimensions. This section revises that work and builds up the terminolog required. Named after René Descartes ( ), a French mathematician. The vertical line is called the -ais. positive -ais nd quadrant negative -ais st quadrant positive -ais The horizontal line is called the -ais. The plane is divided up into four quadrants b etending two perpendicular lines called the coordinate aes and. rd quadrant 4th quadrant The point, where the coordinate aes cross, is called the origin. negative -ais

2 B_Chap0_08-05.qd 5/6/04 0:4 am Page 9 C: Coordinate geometr of straight lines 9 In the diagram below, the point P lies in the first quadrant. Its distance from the -ais is a. Its distance from the -ais is b. We sa that: the -coordinate of P is a the -coordinate of P is b. We write P is the point (a, b) or simpl the point P(a, b). Q is in the second quadrant. The -coordinate is negative and the -coordinate is positive. The point P has Cartesian coordinates (a, b) Q c a P Coordinates of Q are (c, d) d b f h The origin is (0, 0) R e R is the point (e, f ) g S (g, h) The diagram below shows part of the Cartesian grid, where L is the point (, ). The -coordinate of L is and the -coordinate of L is. An point on the -ais has its -coordinate equal to 0. L (, ) M (0, ) J (, 0) K (, 0) 4 4 An point on the -ais has its -coordinate equal to 0. J and K are points on the -ais. N (0, )

3 B_Chap0_08-05.qd 5/6/04 0:4 am Page 0 0 C: Coordinate geometr of straight lines Worked eample. Draw coordinate aes and and plot the points A(, ), B(, ), C(, 4) and D(, 4). The line joining the points C and D crosses the -ais at the point E. Write down the coordinates of E. The line joining the points A and C crosses the -ais at the point F. Write down the coordinates of F. B(, ) 5 4 A(, ) D(, 4) 5 C(, 4) E is on the -ais so its -coordinate is 0. E(0, 4). F is on the -ais so its -coordinate is 0. F(, 0).. The distance between two points The following eample shows how ou can use Pthagoras theorem which ou studied as part of our GCSE course, in order to find the distance between two points. Worked eample. Find the distance AB where A is the point (, ) and B is the point (4, 4). First plot the points, join them with a line and make a right-angled triangle ABC. The distance AC 4. The distance BC 4. Using Pthagoras theorem AB. 9 4 AB From the diagram in Worked eample., ou can see that the distance between the points C(, 4) and D(, 4) is units and the distance between the points A(, ) and C(, 4) is 6 units, but how do ou find the distance between points which do not lie on the same horizontal line or on the same vertical line? 4 A 4 AB.6 (to two significant figures) but since a calculator is not allowed in the eamination for this unit ou should leave our answers in eact forms. Where possible surd answers should be simplified. B C

4 B_Chap0_08-05.qd 5/6/04 0:4 am Page C: Coordinate geometr of straight lines We can generalise the method of the previous eample to find a formula for the distance between the points P(, ) and Q(, ). The distance PR. The distance QR. Q(, ) Using Pthagoras theorem PQ ( ) ( ) PQ [( ) ( ) ]. P(, ) R The distance between the points (, ) and (, ) is ( ) ( ). Worked eample. The point R has coordinates (, ) and the point S has coordinates (5, 6). (a) Find the distance RS. (b) The point T has coordinates (0, 9). Show that RT has length k, where k is an integer. An integer is a whole number. (a) For points R and S: the difference between the -coordinates is 5 () 6 the difference between the -coordinates is 6 8. RS (6) (8) (b) RT () (7) Worked eample.4 The distance MN is 5, where M is the point (4, ) and N is the point (a,a). Find the two possible values of the constant a. If ou draw a diagram ou will probabl find ourself dealing with distances rather than the difference in coordinates in order to avoid negative numbers. This is not wrong. f course ou could have considered the differences as 5 6 and (6) 8 and obtained the same answer. Used a b a b. See section.4. You ma prefer to draw a diagram but it is also good to learn to work algebraicall.

5 B_Chap0_08-05.qd 5/6/04 0:4 am Page C: Coordinate geometr of straight lines The difference between the -coordinates is a 4. The difference between the -coordinates is a () a. MN (a 4) (a ) 5 (a 4) (a ) 5 a 8a 6 4a 8a 4 5 5a 0 a a. Using ( ). Notice that this equation does have two solutions. EXERCISE A Find the lengths of the line segments joining: (a) (0, 0) and (, 4), (b) (, ) and (5, ), The identit (k) (4k) (5k) ma be useful in (j), (k) and (l). (c) (0, 4) and (5, ), (d) (, ) and (, 6), (e) (4, ) and (, 0), (f) (, ) and (6, ), (g) (, 7) and (, ), (h) (, 0) and (6, ), (i) (.5, 0) and (.5, 0) (j) (.5, 4) and (, 6), (k) (8, 0) and (,.5), (l) (.5, ) and (4, 8). Calculate the lengths of the sides of the triangle ABC and hence determine whether or not the triangle is right-angled: (a) A(0, 0) B(0, 6) C(4, ), (b) A(, 0) B(, 8) C(7, 6), (c) A(, ) B(, 4) C(0, 7). From question onwards ou are advised to draw a diagram. In longer questions, the results found in one part often help in the net part. The vertices of a triangle are A(, 5), B(0, ) and C(4, ). B writing each of the lengths of the sides as a multiple of, show that the sum of the lengths of two of the sides is three times the length of the third side. 4 The distance between the two points A(6, p) and B(p, ) is 55. Find the possible values of p. 5 The vertices of a triangle are P(, ), Q(, 0) and R(4, 0). (a) Find the lengths of the sides of triangle PQR. (b) Show that angle QPR 90. (c) The line of smmetr of triangle PQR meets the -ais at point S. Write down the coordinates of S. (d) The point T is such that PQTR is a square. Find the coordinates of T.

6 B_Chap0_08-05.qd 5/6/04 0:4 am Page C: Coordinate geometr of straight lines. The coordinates of the mid-point of a line segment joining two known points From the diagram ou can see that the mid-point of the line segment joining (0, 0) to (6, 0) is (, 0) or or 0 6,0, the mid-point of the line segment joining (0, 0) to (0, 4) is (0, ) or 0,0 (4), and the mid-point of the line segment joining (0, 4) to (6, 0) is (, ) or 0 6, (4 ) 0. Going from P(, ) to Q(5, 8) ou move 4 units horizontall and then 6 verticall. If M is the mid-point of PQ then the journe is halved so to go from P(, ) to M ou move (half of 4) horizontall and then (half of 6) verticall. So M is the point (, ) or M(, 5) or M 5, 8. ( ) A(, ) M, ( ) up B(, ) mid-point 4 5 along mid-point P(, ) M mid-point of line segment is (, ) 4 along up Q(5, 8) Check that M to Q is also along and then up. 6 up ( ) along ( ) Note that ( ) ( ) which is the -coordinate of M. In general, the coordinates of the mid-point of the line segment joining (, ) and (, ) are,.

7 B_Chap0_08-05.qd 5/6/04 0:4 am Page 4 4 C: Coordinate geometr of straight lines Worked eample.5 M is the mid-point of the line segment joining A(, ) and B(, 5). (a) Find the coordinates of M. (b) M is also the mid-point of the line segment CD where C(, 4). Find the coordinates of D. (a) Using,, M is the point, 5 so M(, ). (b) Let D be the point (a, b) then the mid-point of CD is a, b 4 (, ) coordinates of M For this to be true a and b 4 a and b. The coordinates of D are (, ). 5 4 C(, 4) A(, ) B(, 5) M(, ) 4 5 D r, using the idea of journes : C to M is across and down. So M to D is also across and down giving D(, ) or D(, ). EXERCISE B Find the coordinates of the mid-point of the line segments joining: (a) (, ) and (7, ), (b) (, ) and (, ), (c) (0, ) and (6, ), (d) (, ) and (, 6), (e) (4, ) and (, 6), (f) (, ) and (6, ), (g) (, 5) and (, ), (h) (, 5) and (6, ), (i) (.5, 6) and (.5, 0) (j) (.5, ) and (4, ). M is the mid-point of the straight line segment PQ. Find the coordinates of Q for each of the cases: (a) P(, ), M(, 4), (b) P(, ), M(, ), (c) P(, ), M(, 5), (d) P(, 5), M(, 0), (e) P(, 4), M(, ), (f) P(, ), M(, 4 ). The mid-point of AB, where A(, ) and B(4, 5), is also the mid-point of CD, where C(0, ). (a) Find the coordinates of D. (b) Show that AC BD. 4 The mid-point of the line segment joining A(, ) and B(5, ) is D. The point C has coordinates (4, 4). Show that CD is perpendicular to AB.

8 B_Chap0_08-05.qd 5/6/04 0:4 am Page 5 C: Coordinate geometr of straight lines 5.4 The gradient of a straight line joining two known points The gradient of a straight line is a measure of how steep it is. () () () (a) (b) (c) These three lines slope upwards from left to right. The have gradients which are positive. Line () is steeper than line () so the gradient of () is greater than the gradient of (). These three lines are all parallel to the -ais. The are not sloping. Horizontal lines have gradient 0. These three lines slope downwards from left to right. The have gradients which are negative. change in -coordinate The gradient of the line joining two points. change in -coordinate B(, ) ( ) Change in A(, ) ( ) Change in The gradient of the line joining the two points A(, ) and B(, ) is. Lines which are equall steep are parallel. Parallel lines have equal gradients.

9 B_Chap0_08-05.qd 5/6/04 0:4 am Page 6 6 C: Coordinate geometr of straight lines Worked eample.6 (0, 0), P(, 6), Q(0, 5) and R(, ) are four points. (a) Find the gradient of the line segment (i) P, (ii) RQ. (b) Find the gradient of the line segment (i) R, (ii) PQ. Q(0, 5) P(, 6) (c) What can ou deduce from our answers? R(, ) 6 0 (a) (i) Gradient of P ; 0 5 (ii) Gradient of RQ. 0 () 0 (b) (i) Gradient of R 0 ; A diagram is ver helpful. You should tr to position the points roughl in the correct place without plotting the points on graph paper, then ou might see properties such as parallel lines and possible right angles. (ii) Gradient of PQ (c) The lines P and RQ have gradients which are equal so the are parallel. Lines R and PQ are not parallel since their gradients are not equal. So we can deduce that the quadrilateral PQR is a trapezium. EXERCISE C B finding the gradients of the lines AB and CD determine if the lines are parallel. (a) A(, ) B(, 5) C(0, ) D(, ), (b) A(, ) B(5, ) C(4, ) D(, ), (c) A(4, 5) B(4, 5) C(, ) D(0, ), (d) A(6, ) B(, ) C(,0) D(7, ). B finding the gradients of the lines AB and BC show that A(, ), B(, ) and C(6, ) are collinear points. Collinear means in a straight line. A(, ), B(4, ) and C(6, 0) are the vertices of triangle ABC. (a) Find the gradient of each side of the triangle. (b) Which side of the triangle is parallel to P where is the origin and P is the point (, 6)?

10 B_Chap0_08-05.qd 5/6/04 0:4 am Page 7 C: Coordinate geometr of straight lines 7.5 The gradients of perpendicular lines (a) A(, ) A(, ) B(4, ) (b) B(, 4) 4 Rotate the shaded triangles clockwise through 90 as shown, keeping fied. Line A line A Line B line B A(, ) A(, ) B(4, ) B(, 4) Gradient of A Gradient of A Gradient of B 4 Gradient of B 4 A is perpendicular to A Gradient of A Gradient A B is perpendicular to B Gradient of B Gradient B 4 4 P(a, b) In general gradient of P b a gradient of Pa b Lines are perpendicular if the product of their gradients is. P(b, a) Lines with gradients m and m are parallel if m m, are perpendicular if m m. Worked eample.7 Find the gradient of a line which is perpendicular to the line joining A(, ) and B(4, 5).

11 B_Chap0_08-05.qd 5/6/04 0:4 am Page 8 8 C: Coordinate geometr of straight lines Gradient of AB 5 4. Let m be the gradient of an line perpendicular to AB then m m. The gradient of an line perpendicular to AB is. EXERCISE D Write down the gradient of lines perpendicular to a line with gradient: (a) 5, (b), (c) 4, (d), (e) 4. Two vertices of a rectangle ABCD are A(, ) and B(4, ). Find: (a) the gradient of DC, (b) the gradient of BC. A(, ), B(, 6) and C(7, 4) are the three vertices of a triangle. (a) Show that ABC is a right-angled isosceles triangle. (b) D is the point (5, 0). Show that BD is perpendicular to AC. (c) Eplain wh ABCD is a square. (d) Find the area of ABCD..6 The = m + c form of the equation of a straight line Consider an straight line which crosses the -ais at the point A(0, c). We sa that c is the -intercept. Let P(, ) be an other point on the line. If the gradient of the line is m then c m 0 so c m or m c. A(0, c) m 0 P(, ) c m c is the equation of a straight line with gradient m and -intercept c. a b c 0 is the general equation of a line. It has gradient a b and -intercept c. b

12 B_Chap0_08-05.qd 5/6/04 0:4 am Page 9 C: Coordinate geometr of straight lines 9 Worked eample.8 A straight line has gradient and crosses the -ais at the point (0, 4). Write down the equation of the line. The line crosses the -ais at (0, 4) so the -intercept is 4, so c 4. The gradient of the line is, so m. Using m c ou get (4). The equation of the line is 4. Worked eample.9 The general equation of a straight line is A B C 0. Find the gradient of the line, and the -intercept. 4 4 You need to rearrange the equation A B C 0 into the form m c. You can write A B C 0 as B A C or B A B C. Compare with m c, ou see that m B A and c B C, so A B C 0 is the equation of a line with gradient B A and -intercept B C. EXERCISE E Find, in the form a b c 0, the equation of the line which has: (a) gradient and -intercept, (b) gradient and -intercept, (c) gradient and -intercept. Find the gradient and -intercept for the line with equation: (a), (b) 4 5, (c) 4 7, (d) 8, (e) , (f) 0.5 4, (g) 5, (h) 4, (i).5 5, (j) 4.

13 B_Chap0_08-05.qd 5/6/04 0:4 am Page C: Coordinate geometr of straight lines.7 The = m( ) form of the equation of a straight line Consider an line which passes through the known point A(, ) and let P(, ) be an other point on the line. P(, ) If m is the gradient of the line AP, then m or m( ). A(, ) The equation of the straight line which passes through the point (, ) and has gradient m is m( ). Worked eample.0 Find the equation of the straight line which is parallel to the line 4 and passes through the point (, ). The gradient of the line 4 is 4, so the gradient of an line parallel to 4 is also 4. We need the line with gradient 4 and through the point (, ), so its equation is, using m( ), 4( ) or 4 0. Compare 4 with m c Parallel lines have equal gradients. Worked eample. (a) Find a Cartesian equation for the perpendicular bisector of the line joining A(, ) and B(4, 5). (b) This perpendicular bisector cuts the coordinate aes at C and D. Show that CD.5 AB. A(, ) 4 D C 5 B(4, 5)

14 B_Chap0_08-05.qd 5/6/04 0:4 am Page 4 C: Coordinate geometr of straight lines 4 First, we draw a rough sketch. (a) The gradient of AB 5, 4 so the gradient of the perpendicular is. The mid-point of AB is 4, (5) (, ). The perpendicular bisector is a straight line which passes through the point (, ) and has gradient, so its equation is () ( ) or. m m Using,. Using m( ). (b) Let C be the point where the line cuts the -ais. When 0, 0, so we have C(0, ). Let D be the point where the line cuts the -ais. When 0, 0 9, so we have D(9, 0) Distance CD (9 0) (0 ()) Distance AB (4 ) (5 ) C D AB so CD.5 AB An point on the -ais has -coordinate 0. An point on the -ais has -coordinate 0. Using ( ) ( ). Using a b a. b See section.4. EXERCISE F Find an equation for the straight line with gradient and which passes through the point (, 6). Find a Cartesian equation for the straight line which has gradient and which passes through (6, 0). Find an equation of the straight line passing through (, ) which is parallel to the line with equation 4. 4 Find an equation of the straight line that is parallel to 4 0 and which passes through (, ). Because a straight line equation can be arranged in a variet of was ou are usuall asked to find an equation rather than the equation. All correct equivalents would score full marks unless ou have been asked specificall for a particular form of the equation.

15 B_Chap0_08-05.qd 5/6/04 0:4 am Page 4 4 C: Coordinate geometr of straight lines 5 Find an equation of the straight line which passes through the origin and is perpendicular to the line. 6 Find the -intercept of the straight line which passes through the point (, ) and is perpendicular to the line. 7 Find a Cartesian equation for the perpendicular bisector of the line joining A(, ) and B(0, 6). 8 The verte, A, of a rectangle ABCD, has coordinates (, ). The equation of BC is. Find, in the form m c, the equation of: (a) AD, (b) AB. 9 Given the points A(0, ), B(5, 4), C(4, ) and E(, ): (a) show that BE is the perpendicular bisector of AC, (b) find the coordinates of the point D so that ABCD is a rhombus, (c) find an equation for the straight line through D and A. Hint for (b). The diagonals of a rhombus are perpendicular and bisect each other. 0 The perpendicular bisector of the line joining A(0, ) and C(4, 7) intersects the -ais at B and the -ais at D. Find the area of the quadrilateral ABCD. Show that the equation of an line parallel to a b c 0 is of the form a b k 0..8 The equation of a straight line passing through two given points The equation of the straight line which passes through the points (, ) and (, ) is B(, ) P(, ) The derivation of this equation is similar to previous ones and is left as an eercise to the reader. A(, ) Worked eample. Find a Cartesian equation of the straight line which passes through the points (, ) and (, 0).

16 B_Chap0_08-05.qd 5/6/04 0:4 am Page 4 C: Coordinate geometr of straight lines 4 If ou take (, ) (, ) and (, ) = (, 0) and substitute into the general equation, ou get, 0 or which leads to. Note that ou can check that our line passes through the points (, ) and (, 0) b seeing if the points satisf the equation. Checking for (, ) LHS RHS Checking for (, 0) LHS 0 RHS 0.9 The coordinates of the point of intersection of two lines In this section ou will need to solve simultaneous equations. If ou need further practice, see section. of chapter. 4 A point lies on a line if the coordinates of the point satisf the equation of the line. When two lines intersect, the point of intersection lies on both lines. The coordinates of the point of intersection must satisf the equations of both lines. The equations of the lines must be satisfied simultaneousl. Given accuratel drawn graphs of the two intersecting straight lines with equations a b c 0 and A B C 0, the coordinates of the point of intersection can be read off. These coordinates give the solution of the simultaneous equations a b c 0 and A B C (, ) 4 0 From the diagram, ou can clearl read off (, ) as the point of intersection of the lines 0 and 4 0. So the solution to the simultaneous equations 0 and 4 0 is,. To find the coordinates of the point of intersection of the two lines with equations a b c 0 and A B C 0, ou solve the two linear equations simultaneousl. (, )

17 B_Chap0_08-05.qd 5/6/04 0:4 am Page C: Coordinate geometr of straight lines Worked eample. Find the coordinates of the point of intersection of the straight lines with equations and 4. P At the point of intersection P, and 4. So eliminating gives 4. Rearranging gives 4. The -coordinate of P is. To find the -coordinate of P, put into The point of intersection is (, ). Worked eample.4 The straight lines with equations 5 7 and 7 intersect at the point R. Find the coordinates of R. To find the point of intersection R, ou need to solve the simultaneous equations 5 7 [A] 7 [B] 5 49 [C] 9 9 [D] Substituting in [A] gives 0 7. The coordinates of the point R are (, ). Hint. Checking that the coordinates satisf each line equation is advisable, especiall if the result is being used in later parts of an eamination question. It can usuall be done in our head rather than on paper. Checking that (, ) lies on the line ; LHS RHS Checking that (, ) lies on the line 4 ; LHS RHS 4 R Multipl equation [A] b 7 and equation [B] b. Adding [C] [D]. Checking in [A] LHS0()7RHS ; and in [B] LHS6 7()

18 B_Chap0_08-05.qd 5/6/04 0:4 am Page 45 C: Coordinate geometr of straight lines 45 EXERCISE G Verif that (, 5) lies on the line with equation. Which of the following points lie on the line with equation 6: (a) (, 0), (b) (, 0), (c) (4, ) (d) (, 6), (e) (0, )? Find the coordinates of the points where the following lines intersect the -ais: (a) 4, (b) 6, (c) 6 0, (d) The point (k, k) lies on the line with equation 6 0. Find the value of k. 5 Show that the point (4, 8) lies on the line passing through the points (, ) and (7, ). 6 (a) Find the equation of the line AB where A is the point (, 7) and B is the point (5, ). (b) The point (k, ) lies on the line AB. Find the value of the constant k. 7 A(5, ), B(, ), C(, ) and D(4, ) are the vertices of the quadrilateral ABCD. (a) Find the equation of the diagonal BD. (b) Determine whether or not the mid-point of AC lies on the diagonal BD. 8 Find the coordinates of the point of intersection of these pairs of straight lines: (a) 7 and, (b) 7 and, (c) 5 6 and 8, (d) 8 and 40, (e) 7 and 5, (f) and 5 9, (g) and 5 6, (h) 8 and Point A has coordinates,, point B has coordinates, 6 60 and point C has coordinates 9, 6 5. The straight line AC has equation 65 0 and the straight line BC has equation Write down the solution of the simultaneous equations 65 0 and

19 B_Chap0_08-05.qd 5/6/04 0:4 am Page C: Coordinate geometr of straight lines 0 The straight line passes through the point of intersection of the two straight lines and 7 5. Write down the solution of the simultaneous equations and 7 5. Worked eample.5 ABCD is a parallelogram in which the coordinates of A, B and C are (, ), (7, ) and (, ), respectivel. (a) Find the equations of AD and CD. (b) Find the coordinates of D. (c) Prove that angle BAC 90. (d) Calculate the area of the parallelogram. (e) Show that the length of the perpendicular from A to BC 665 is. D C(, ) A(, ) B(7, ) Alwas start with a good sketch; this helps to spot obvious errors like wrong signs for gradients or wrong quadrants for points. Note that when it sas the parallelogram ABCD, the points must be connected in that order which determines where D must be. (a) AD is parallel to BC and CD is parallel to BA. Gradient of BC () 7 () 8 gradient of AD = 8 AD is a line through (, ) and has gradient 8 equation is ( ) or Gradient of BA 7 6 Gradient of BA gradient of CD. so its pposite sides of a parallelogram are equal and parallel. Using. Using m( ). You could check the signs of the two gradients using our sketch. CD is a line through (, ) and has gradient so its equation is () [ ()] or 5.

20 B_Chap0_08-05.qd 5/6/04 0:4 am Page 47 C: Coordinate geometr of straight lines 47 (b) D is the point of intersection of AD and CD. Solving 8 5 and 5 simultaneousl, adding gives Substitution in the second equation gives () 5 7, i.e. D(7, ) (c) Gradient of AC ( ) 4 ( ). From earlier work, gradient of BA. Gradient of AC gradient of BA, so AC is perpendicular to BA and angle BAC 90. (d) Since angle BAC 90 ou can use: Area of parallelogram ABCD base height AB AC. AB (7 ) ( ) AC ( ) ( ) So area of parallelogram ABCD = 0 square units (e) Using the base of the parallelogram as CB and letting h length of the perpendicular from A to BC, area of parallelogram ABCD CB h, 0 (7 ) ( ( ) ()) h, h, h Alternativel, since BA is equal and parallel to CD, the journe from B A (6 left then up) is the same as from C to D. D( 6, ) or D(7, ). From the sketch ou know D is in the nd quadrant. Checking in both equations 8 (7) 5 (7) 5 Angle BAC 90 if AC and BA are perpendicular. Aim to show that m m. Hint. In eam questions look for possible links between the parts. ( ) ( ) Keep our answers in surd form. Hint. In eam questions look for links between the parts: Length of perpendicular from A to BC is required. We have just found the area of the parallelogram ABCD. Can the two be linked? MIXED EXERCISE The point A has coordinates (, ) and is the origin. (a) Write down the gradient of A and hence find the equation of the line A. (b) Show that the line which has equation 4 6 : (i) is perpendicular to A, (ii) passes through the mid-point of A. [A]

21 B_Chap0_08-05.qd 5/6/04 0:4 am Page C: Coordinate geometr of straight lines The line AB has equation 5 7. The point A has coordinates (, ) and the point B has coordinates (, k). (a) (i) Find the value of k. (ii) Find the gradient of AB. (b) Find an equation for the line through A which is perpendicular to AB. (c) The point C has coordinates (6, ). Show that AC has length p, stating the value of p. [A] The point P has coordinates (, 0) and the point Q has coordinates (4, 4). (a) Show that the length of PQ is 5. (b) (i) Find the equation of the perpendicular bisector of PQ. (ii) This perpendicular bisector intersects the -ais at the point A. Find the coordinates of A. 4 The point A has coordinates (, 5) and the point B has coordinates (, ). (a) (i) Find the gradient of AB. (ii) Show that the equation of the line AB can be written in the form r s, where r and s are positive integers. (b) The mid-point of AB is M and the line MC is perpendicular to AB. (i) Find the coordinates of M. (ii) Find the gradient of the line MC. (iii) Given that C has coordinates (5, p), find the value of the constant p. [A] 5 The points A and B have coordinates (, 5) and (9, ), respectivel. (a) (i) Find the gradient of AB. (ii) Find an equation for the line AB. (b) The point C has coordinates (, ) and the point X lies on AB so that XC is perpendicular to AB. (i) Show that the equation of the line XC can be written in the form 4 7. (ii) Calculate the coordinates of X. [A] 6 The equation of the line AB is 5 6. (a) Find the gradient of AB. (b) The point A has coordinates (4, ) and a point C has coordinates (6, 4). (i) Prove that AC is perpendicular to AB. (ii) Find an equation for the line AC, epressing our answer in the form p q r, where p, q and r are integers. (c) The line with equation also passes through the point B. Find the coordinates of B. [A]

22 B_Chap0_08-05.qd 5/6/04 0:4 am Page 49 C: Coordinate geometr of straight lines 49 7 The points A, B and C have coordinates (, 7), (5, 5) and (7, 9), respectivel. (a) Show that AB and BC are perpendicular. (b) Find an equation for the line BC. (c) The equation of the line AC is 0 and M is the mid-point of AB. (i) Find an equation of the line through M parallel to AC. (ii) This line intersects BC at the point T. Find the coordinates of T. [A] 8 The point A has coordinates (, 5), B is the point (5, ) and is the origin. (a) Find, in the form m c, the equation of the perpendicular bisector of the line segment AB. (b) This perpendicular bisector cuts the -ais at P and the -ais at Q. (i) Show that the line segment BP is parallel to the -ais. (ii) Find the area of triangle PQ. 9 The points A(, ) and C(5, ) are opposite vertices of a parallelogram ABCD. The verte B lies on the line 5. The side AB is parallel to the line 4 8. Find: (a) the equation of the side AB, (b) the coordinates of B, (c) the equations of the sides AD and CD, (d) the coordinates of D. 0 ABCD is a rectangle in which the coordinates of A and C are (0, 4) and (, ), respectivel, and the gradient of the side AB is 5. (a) Find the equations of the sides AB and BC. (b) Show that the coordinates of B is (, ). (c) Calculate the area of the rectangle. (d) Find the coordinates of the point on the -ais which is equidistant from points A and D. [A]

23 B_Chap0_08-05.qd 5/6/04 0:4 am Page C: Coordinate geometr of straight lines Ke point summar The distance between the points (, ) and (, ) p is ( ) ( ). The coordinates of the mid-point of the line segment p joining (, ) and (, ) are,. The gradient of a line joining the two points A(, ) p5 and B(, ). 4 Lines with gradients m and m : p7 are parallel if m m, are perpendicular if m m. 5 m c is the equation of a straight line with p8 gradient m and -intercept c. 6 a b c 0 is the general equation of a line. p8 It has gradient a b and -intercept c. b 7 The equation of the straight line which passes p40 through the point (, ) and has gradient m is m( ). 8 The equation of the straight line which passes p4 through the points (, ) and (, ) is. 9 A point lies on a line if the coordinates of the point p4 satisf the equation of the line. 0 Given accuratel drawn graphs of the two intersecting p4 straight lines with equations a b c 0 and A B C 0, the coordinates of the point of intersection can be read off. These coordinates give the solution of the simultaneous equations a b c 0 and A B C 0. To find the coordinates of the point of intersection of p4 the two lines with equations a b c 0 and A B C 0, ou solve the two equations simultaneousl.

24 B_Chap0_08-05.qd 5/6/04 0:4 am Page 5 C: Coordinate geometr of straight lines 5 Test ourself What to review Calculate the distance between the points (, ) and (7, 9). Section. State the coordinates of the mid-point of the line segment PQ Section. where P(, ) and Q(7, ). Find the gradient of the line joining the points A and B Section.4 where A is the point (, ) and B is the point (5, 4). 4 The lines CD and EF are perpendicular with points C(, ), Section.5 D(, 4), E(, 5) and F(k, 4). Find the value of the constant k. 5 Find a Cartesian equation of the line which passes through Sections.5 and.7 the point (, ) and is perpendicular to the line Find the point of intersection of the lines with equations Section.9 5 and 4 9. Test ourself ANSWERS. (5, ).. 4 k (, ).

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