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- 1. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 60CHAPTER 6- COORDINATE GEOMETRY6.1 DISTANCE BETWEEN TWO POINTSyy2 Q (x2, y2)y2- y1y1 P(x1, y1)x2- x1xTo find the distance or length of PQ, use the concept of Pythagoras’ Theorem.2122122)()( yyxxPQ −+−=Therefore:where D is distance.Example 1:The distance point A(6, 3t) and point B(12, -t) is 10 units. Find the possible values of t.Solution:2464161636100)4(6100)3()612(102222222±===+=−+=−−+−=tttttttExample 2:Point A(h, 2h) and point B(h -3 , 2h + 1) are two points which are equidistant from the origin. Find thevalue of h.Solution:2222)012()03()02()0( −++−−=−+− hhhh212212 )()( yyxxD −+−=How to obtain the formula?Square the both sides
- 2. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 61510210255144964)12()3()2(2222222222==+−=++++−=+++−=+hhhhhhhhhhhhhhhEXERCISE 6.11. Find the perimeter of triangle ABC with points A(2, 1), B(4, 5) and C(-2, 8).2. The points (-3, -5) lies on the circumference of a circle with centre (2, 1). Calculate the radius of thecircle.3. Given that the distance between points A(1,3) and B(7, k) is 10 units . Find the possible value of k.4. Given point A( 1, 7) and B(p, 2) and the distance between the points A and B is 13 units. Find the valueof p.6.2 DIVISION OF A LINE SEGMENTLine segment is a line that has distant.6.2.1 Mid-pointThe formula to find mid-points is just the same as we have learned in Form Two that is:Example:Given C(2, 5) is the mid-point of the point B(h, 3) and point D(-4, k). Find the values of h and k.Solution:Use the formula,844242==−+−=hhh7103235==++=kkkEXERCISE 6.2.11. The coordinates of A and B are (m, 5) and (6, n) respectively. Find the values of m and n if the mid-point of the points is (4, 10).2. The coordinates of M and N are (4, 2) and (6, 5) respectively. Find the mid-point of these two points.3. The coordinates of X and Y are (-1, b) and (a, 7) respectively. Find the values of a and b if the mid-point of the points is (1, 2).)2,2(),( 2121 yyxxyx++=
- 3. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 626.2.2 Point which divides a line segment in the ratio m: nQ (x2, y2)ny2- y y2- yX(x, y)m y- y1y- y1P(x1, y1) x- x1 x2- xIf the line PQ moves downwards, it will reach the horizontal line.From that we know that:nmmxnxxmxnxnmxmxnxmxnxmxmxnxnxxxmxxnnmxxxx++=+=++=+−=−−=−=−−212121212121)()()(If the line PQ moves upwards, it will reach the vertical linenmmynyymynynmymynymynymymynynyyymyynnmyyyy++=+=++=+−=−−=−=−−212121212121)()()(Hence, the formula to find the point that divides the line segment in ratio m : n isWhen nm = , the point will be the midpoint of the line segment.++++=nmmynynmmxnxyx 2121,),(
- 4. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 63Example:Find the coordinates of point P that divides the straight line that joins E(-6, 10) and F(4, -5) in the ratio2: 3E(-6, 10)Solution:m: n = 2 : 3 2P(x, y)3F(4, -5)EXERCISE 6.2.21. Point R divides the line segment joining J(-1, -7) and Q(10 ,7) internally in the ration PR: RQ = 1: 3. Findthe coordinate of R.2. The points P(t, 2t), Q(2a, b) and R(4a, 3b) are on a straight line. Q divides PR internally in the ratio 1: 4.Show that ab 6= .3. Given points A(k, 5), B(0, 3) and C(5, 4). Find the possible values of k if the length of AB is twice thelength of BC.6.3 AREA OF POLYGONS6.3.1 Area Of TriangleC(x3, y3)B(x2, y2)A(x1, y1)K L MThe area of ABC∆ = Area of trapezium ACLK + Area of trapezium BCLM - Area of trapezium ABMKooo+−+++−=32)5)(2()10)(3(,32)4)(2()6)(3(),( yxP++++=nmmynynmmxnxyx 2121,),( −+−=51030,5818),( yxP( )4,2−=How to obtain the formula?
- 5. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 64))((21))((21))((21122132321313 xxyyxxyyxxyy −+−−++−+=[ ])()(21)(21)[21)([21312312133221312312133221211112223323322231111333211112223323322231111333yxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyx++−++=−−−++=++−−−−++−−+=−−+−−−++−−+=Area of ∆ ABC =1133221121yxyxyxyx=[ ])()(21312312133221 yxyxyxyxyxyx ++−++=The formula of the area of triangle is6.3.2 Area of QuadrilateralThe formula of the area of quadrilateral is( ) ( )413423121443322121yxyxyxyxyxyxyxyx +++−+++=Example 1:Find the value of m if the point P(m, 2), Q(4, -3) and R(-2, 5) lie on a straight line.Solution:Method 1- Using Concept Area Of Triangle025234221=−−mm1133221121yxyxyxyxThe area of a straight line iszero.114433221121yxyxyxyxyxSimplify
- 6. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 6502805141630)514(1630)568()4203(=+−=−−+−=+−+−=++−−+−mmmmmmm028028=+−=+−mm4128==mmMethod 2Using the concept of gradient of straight line68)2(453−=−−−−=QRQRmm34−=411415164344)3(2==−=−−=−−−=mmmmmm QRPQExample 2:Find the possible values of k if the area of triangle with vertices A (9, 2), B(4, 12) and C(k, 6) is 30 unit2Solution:302961242921=k60107060)54128()224108(=−=++−++kkkPoint P, Q and R lie on thesame line, so QRPQ mm =
- 7. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 66(i) 601070 −=− k11010==kk(ii) 601070 −=− k1313010==kkExample 3:Find the area of quadrilateral KLMN given K (1, 3), L(-1, 2) and M(-4, -3) and N(6, -9).Solution:319634213121−−−−=25.489721385321)38(5921)91883()183632(21unit==+=−−=−−−−−+++=EXERCISE 6.31. Find the possible values of p if the area of triangle with the vertices D(p, -p), E(1, 0) and F(-3, 6) is 10unit2.2. Find the area of the rhombus PQRS if the coordinates of the points P, Q and R are (6, 4). (8, 7) and(-6, 3) respectively.3. The points G(4, -2), F(1, 1) and H(-2, p) lie on a straight line. Find the value of p.4. Find the area of the triangle PQR if the coordinates of the vertices are:(a) P(1, 3), Q (4, 2) and R (7, 0)(b) P(-2, 5), Q (7, -2) and R (-3, 1)(c) P(-1, -5),Q (1, -4) and R(-1, -3)The modulus sign will always result apositive number. If k1070 − wouldresult 60, there will be two values of k. Ifthe value of k is 1, the value in themodulus is 60. If the value of k is 13, thevalue in the modulus is -60 and at lastbecomes 60 because the modulus signwill always result a positive number.114433221121yxyxyxyxyx
- 8. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 676.4 EQUATION OF A STRAIGHT LINE6.4.1 x-intercept and y-Intercept of a straight lineybx0 a1. The line intersects with the x-axis at a and the line intersects with the y-axis at b2. a is called x-intercept and b is called y-intercept.EXERCISE 6.4.11. Determine the x-intercept for the following straight lines:(a) 132=+yx(b) 0943 =−− yx(c) 66 −= xy2. Determine the y-intercept for the following straight lines:(a) 151=+yx(b) 01475 =−− yx(c) 653 −= xy6.4.2 The Gradient of a Straight Line(i) (ii) (iii) (iv)The gradient is positive the gradient is negative the gradient is undefined the gradient is zero
- 9. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 68Finding the gradient (m) of a straight line1-Example:yA(0,5)x0 B(0, 4)1212xxyym−−=4005−−=ABm45−=2-y(0, b)x0 (a, 0)abm−−=001212xxyym−−=
- 10. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 69ab−=Hence,Example:yA(0,5)x0 B(0, 4)erceptxerceptymintint−−=45−=ABm3-Example: yA(0,5)5x0 4 B(0, 4)θerceptxerceptymintint−−−=θtan=m
- 11. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 70θtan=ABm45−=ABmEXERCISE 6.4.2Find the gradient of the following points.(a) A (2, 3) and B (4, 5)(b) M (-3, 1) and N (4, -2)(c) P (2, 3) and Q (4, 3)(d) C (2, 5) and D (2, 8)(e) F(0, 6) and G( 3, 0)6.4.3 The Equation of a Straight Line1-General FormThe equation of general form is 0=++ cbyaxExample:Given the equation of a straight line is 542 += xy . Change the equation into the general form.Solution:542 += xy0524 =+− yx2-Gradient FormThe equation of gradient form is cmxy += where m is the gradient and c is y-interceptExample:Given the equation of a straight line is 542 += xy . Determine the gradient and the y-intercept ofthe straight line.Solution:542 += xy252 += xyHence, the gradient of the straight line is 2 while and the y-intercept of the straight line is25.
- 12. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 713-Intercept formThe equation of intercept form is 1=+byaxwhere a is x-intercept and b is y-intercept.Example:Given the equation of a straight line is 632 += xy . Convert the equation into the intercept form.Hence, state the x-intercept and y-intercept of the straight line.Solution:632 += xy13213216263623=+−=+−=+−=+−yxyxyxyxHence, the x-intercept of the straight line is -2 and y-intercept of the straight line is 3.Example:Find the equation of the straight line which has a gradient of -3 and passes through the mid-point of theline joining A (1, 4) and B(7, -2).Solution:Mid-point of AB −++=2)2(4,217( )1,4=The equation of the straight line which has a gradient of -3 and passes through (4, -1) isxyxy3121341−=−−=−−0133 =−+ yx or 133 +−= xyEXERCISE 6.4.31. Write each of the following equations to intercept form. Hence, state the gradient of the straight line.(a) 042 =−+ yx(b) 63 =− yx(c) 234 =+ yx(d) 132 =+ yx2. Find the equation of the straight line which has a gradient of -2 and passes through point B (7, -2).Use the general point (x, y) andspecific point (4, -1) to find thegradient of the line and it is equalto the given gradient.
- 13. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 726.4.4 The Point of intersection of two Straight lines1. When two lines intersect, the point of intersection lies on both lines.2. This means the coordinates of the point satisfy both the equations of the lines.3. Therefore, we need to solve the equation simultaneously in order to determine the point ofintersection.Example 1:The straight line which has a gradient of 2 and passes through the point (4, -1) intersects with thestraight line 6−=+ yx at the point P. Find the coordinates of the point P.Solution:First of all, we have to find the equation of the straight line.9282124)1(=−−=+=−−−yxxyxy6−=+ yx+ ,133==xxSubstitute 1=x into ,79)1(2−==−yyHence the coordinates of P is (1, -7)Example 2:The straight line 04 =−+ yx and 01132 =−+ yx intersect at point A. Find the equation of thestraight line which passes through the point A and point B (5, 2).Solution:082204=−+=−+yxyx01132 =−+ yx- ,303==−yy121 21122 1
- 14. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 73Substitute 3=y into ,12208)3(22===−+xxxHence the coordinates of A is (1, 3)1532−−=ABm41−=ABmThe equation of the straight line that passes through point A and B isxyxy−=−−=−−112441130134 =−+ yx or 134 =+ yx or41341+−= xyEXERCISE 6.4.41. Find the coordinates of the point of intersection of the line 01134 =−+ yx and 01762 =+− yx .2. Find the points of intersection of the following pairs of straight lines.(a) 025 =−− yx (b) 42 += xy052 =+− yx 5+= xy3. Find the equation of that is parallel to the line 52 += xy and passing through the point ofintersection of lines 092 =−− yx and 22 =+ yx .6.5 PARALLEL AND PERPENDICULAR LINESParallel Lines1. When two lines are parallel, they have the same gradient.B QA P2. If line AB and line PQ are parallel, so PQAB mm = .1We can write the equation in anyform. Either the general form orintercept form or gradient form.
- 15. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 74Example:The straight line AB passes through the point (6, 3) is parallel to the straight line PQ. Given point P (0, 2)and point Q(4, 0). Find the equation of the straight line AB.Solution:First of all, we have to find the gradient of straight line PQ42−=PQm21−=PQAB mm =21−=ABmThe equation of the straight line that passes through point (6, 3) isxyxy−=−−=−−66221630122 =−+ yx or 122 =+ yx or 621+−= xyPerpendicular Linesyx1. Given that line AB and BC are perpendicular to each other.2. We already know θtan=m .θtan1 =m αtan2 =mABBC=BCAB−=BCAθ α1m2mWe can write the equation in anyform. Either the general form orintercept form or gradient form.
- 16. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 753. WhenBCABABBCmm −×=× 21Hence, if two lines are perpendicular to each other, then the product of their gradient is 1− .Example:Given the straight line 9−= txy and 32 += xy is perpendicular to each other. Find the value of t.Solution:32 += xy21 =m9−= txytm =2121 −=× mm211212−=−=−=×tttEXERCISE 6.51. The equation of the straight line PQ is 0786 =+− xx . Each of the following straight line is parallel toPQ. Find the value of t in each case.(a) 064 =−+ ytx(b) 82+= xty(c) 012 =−− tyx2. Find the equation of the straight line which passes through point B (2, -5) and perpendicular to thestraight line 13 +−= xy .3. PQRS is a rhombus with P (0, 5) and the equation of QS is 12 += xy . Find the equation of diagonal ofPR.4. Find the value of h if the straight line 02 =+− hxy is perpendicular to the straight line035 =++ xy .5. Given that the equation of the line PQ is 1532 += xy and point Q lies on the y-axis. Point R is (4, 1)lies on line QR. Find the equation of QR if the line PQ and QR are perpendicular to each of other.Use the concept121 −=× mm
- 17. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 766.6 LOCUS OF A MOVING POINTLocus represents the path followed by a moving point with the reference to one or more fixed points,satisfying certain conditions.6.6.1 Equation of LocusEquidistant from a fixed pointyP (x, y)3 unitA (1, 1)xThe equation of locus is0722912129)1()1(3)1()1(22222222=−−−+=+−++−=−+−=−+−yxyxyyxxyxyxEquidistant from two fixed pointsFind the equation of the locus of a moving point P such that its distance from the point A (1, 2) and pointB (3, 4) are equal.B (3, 4)A (1, 2)Equation of locus is actuallyinvolving the distance betweentwo points. So we have to usethe formula of the distancebetween two points to find theequation of locus. There is nospecific formula to find theformula to find the equation oflocus.In this case, P is the moving pointsuch that its distance is always 3unit from point A.05020442586542168964412)4()3()2()1()4()3()2()1(2222222222222222=−+=−++−−+=+−−++−++−=+−++−−+−=−+−−+−=−+−=yxyxyxyxyxyxyyxxyyxxyxyxyxyxBPAPlocus),( yxPSquare the both sidesSquare the both sides
- 18. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 77Constant ratio between two fixed pointsFind the equation of locus of a moving point R such that its distance from the point E (4, 3) and the pointF (1, 5) is in the ration 3: 1.020984108896168225909918996168)251012(9)3()4()5()1(331322222222222222=+−−++−++−=+−++−+−++−=+−++−−+−=−+−==yxyxyyxxyyxxyyxxyyxxyxyxRERFRFREEXERCISE 6.61. Given the point A (0, 3) and the point B (1, 4).Find the equation of locus of a moving point Q such thatAQ= 2QB.2. Given A (5, -2) and B (2, 1) are two fixed points. Point Q moves such that the ratio of AQ: QB = 2: 1.Show that the equation of the locus of point Q is 034222=−−−+ yxyx .3. P is a moving point such that its distances from the points A(2, 5) and B(0, 3) is in the ratio of 2: 1. Findthe equation of locus P.4. N is a locus which moves in such a way that NP=NQ. Given that P and Q are coordinates (-3, 6) and(6,- 4) respectively, find the equation of locus N.5. Show that the equation of the locus of a point that moves in such way that is distance from a fixedpoint (3, -1) is 6 units, is by 0262622=−+−+ yxyx .CHAPTER REVIEW EXERCISE1. Given the equation of straight lines AB and CD are 16=+kyxand 0432 =−+ yx respectively, findthe value of k if AB is perpendicular to CD.2. The coordinates of the point A and B are (-2, 3) and (7, -3) respectively. Find(a) the coordinates of C given that AB: BC = 1: 2.(b) the equation of the straight line that passes through B and is perpendicular to AB.3. ABCD is a parallelogram with coordinates A (-2, 3), B (3,4), C (2, -1) and D (h, k).(a) Find the value of h and k.(b) Find the equations of the diagonals AC and BD.(c) State the angle between the diagonals AC and BD.(d) Find the area of the parallelogram ABCD.Square the both sides
- 19. Additional Mathematics Module Form 4Chapter 6- Coordinate Geometry SMK Agama Arau, PerlisPage | 784. P, Q and R are three points on a straight line. The coordinates of P and R are (-2, 3) and (3, 5)respectively. Point Q lies on the y-axis. Find(a) the ratio PQ: QR(b) the coordinates of point Q5. H is a point which moves such that its distance from point P (1, -2) and Q (-3, 4) is always equal. Showthat the equation of the locus H is given by the equation 0532 =+− yx .6. Find the equation of straight line that passes through point P( 1 -2) and parallel to 4x – 2y = 8.7. In diagram below, °=∠ 90PRS .Find(a) the coordinates of R(b) the ratio of PQ: QR(c) the equation of RS.8. Diagram below shows a triangle BCD. The point A lies on the straight line BD.Find(a) the value of k(b) the equation of CD, giving your answer in general form.)9,6(−PSQR3xy0)7,1(−C)8,13(A)2,5(B),17( kDx0y

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