4038 ADDITIONAL MATHEMATICS TOPIC 2: GEOMETRY AND TRIGONOMETRY SUB-TOPIC 2.2 COORDINATE GEOMETRY IN TWO DIMENSIONS

Size: px

Start display at page:

Download "4038 ADDITIONAL MATHEMATICS TOPIC 2: GEOMETRY AND TRIGONOMETRY SUB-TOPIC 2.2 COORDINATE GEOMETRY IN TWO DIMENSIONS"

Frederick Golden
5 years ago
Views:

1 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC. COORDINATE GEOMETRY IN TWO DIMENSIONS CONTENT OUTLINE. Condition for two lines to be parallel or perpendicular. Mid-point of line segment 3. Finding the area of rectilinear figure given its vertices 4. Graphs of equations a. y a n, where n is a simple rational number b. y k 5. Coordinate geometry of the circle with the equation ( a) + (y b) r and + y + g + fy + c 0 6. Transformation of given relationships, including y a n and y kb, to linear form to determine the unknown constants from the straight line graph Eclude:. Finding the equation of the circle passing through three given points.. Intersection of two circles

4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS A Introduction In Mathematics, we have learnt the above few.

2 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS A Introduction In Mathematics, we have learnt the above few. Let us revise: - The Length of AB, where A (, y ) and B (, y ), has a length of - + y - y - The gradient of a line refers to the vertical change as we move unit across the ais. Hence, Gradient of line L Change in y coordinates Change in coordinates y -y - Whereby (, y ) and (, y ) are two points on the same line L. - The general equation of a line segment is y m + c, whereby m is the gradient of the line and c is the y-intercept of the line. For Additional Mathematics, you will require the knowledge of these above topics and more. B Gradient of a Line Parallel Lines Notice that all these lines move up units and across 3 units (hence the gradient is 3 ). In general, lines with same gradient are parallel and vice versa. Perpendicular Lines Look at the two perpendicular lines on the right. Gradient of line A - a Gradient of line B a Product of gradients ( a )( - a ) -. In general, if the product of the gradients of two lines is -, they are perpendicular. Also, the gradient of a line (m ) is the negative reciprocal of the other line (- m ). Horizontal Lines As you have learnt in Mathematics, horizontal lines have a gradient of 0 as they continue to stretch to the right but do not have any change in y coordinates. Vertical Lines As you have learnt in Mathematics, vertical lines have an undefined gradient as they continue to rise upwards but do not have any change in coordinates.

3 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS C Equation of a Straight Line If three points are on the same line, they are collinear. Collinear points can be identified by the fact that the gradients of any two points on the line are equal. Consider the collinear points A (, y ), B (, y ) and C (, y). Knowing that the gradients are equal, we gather that: Gradient of AB Gradient of AC y - y y - y - - y - y y y - - We thus gather that the equation of a straight line is y y m( ), where m is the gradient y - y - of the line and (, y ) are the coordinates of any given point on the line. D Midpoint of a Line Segment Let us find the coordinates of midpoint M of the line joining A (, y ) and B (, y ). From A to B, we move units across and y y units up. Hence, to reach M from A, we move - units up. units across and y - y We hence derive that the coordinates of M are ( + -, y + y - y ) units, which is also equal to ( +, y + y ). In general, the midpoint of a line connecting points (, y ) and (, y ) is: ( +, y + y ).

B (, y) y y3 Let s draw a few lines down from points A, B and C to the -ais. y 3 3 The points at which they touch the -ais are J, K and L respectively.

4 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS E Area of Rectilinear Figures A (, y) C (3, y3) Consider the following triangle ABC with points A (, y ), B (, y ) and C ( 3, y 3 ). B (, y) y y3 Let s draw a few lines down from points A, B and C to the -ais. y 3 3 The points at which they touch the -ais are J, K and L respectively. To find its area, we have to manipulate the areas of the inscribed trapeziums. That is to say: + - Area of Trapezium ABKJ ( )(y + y )( ) Area of Trapezium CAJL ( )(y + y 3 )( 3 ) Area of Trapezium CBKL ( )(y + y 3 )( 3 ) ( )( y y + y y ) + ( )( 3y y + 3 y 3 y 3 ) - ( )( 3y y + 3 y 3 y 3 ) ( )( y y + y y + 3 y y + 3 y 3 y 3-3 y + y - 3 y 3 + y 3 ) ( )( y y + y y + 3 y y + 3 y 3 y 3-3 y + y - 3 y 3 + y 3 ) ( )( y + y + 3 y y 3-3 y + y 3 ) ( )( y + + y y y y 3 3 y ) Difficult to remember? We can rearrange this above complicated formula to become: 3 y y y 3 y ( )[(sum of products) (sum of products)] This formula can also be etended to find the areas of polygons with a specified number of sides, given the coordinates/ vertices. For a figure with n sides, the formula for its area would be re-arranged like this: 3 y y y 3 n y n y ( )[(sum of products) (sum of products)] When applying this formula, you must take note of a few things: - The vertices must always be arranged in an anti-clockwise manner. - Any point can be used as the starting, and you must remember to place the point at the end as well.

5 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS F Worked Eamples on Sections B, C, D, E A line is drawn through the point A(4, 6) parallel to the line y, meets the y-ais at the point B. A line drawn through A(4, 6) perpendicular to AB, meets the line y at the point C. (i) Calculate the coordinates of B. When the line meets the y-ais, 0. Hence, let B (0, B y ) y - y - y Since AB parallel to y, then gradient of AB, and equation of AB: y 6 ( 4) Y + 6 Y + 4 Sub in B: B y (0) Hence, B (0, 4) (ii) Calculate the coordinates of C. Gradient of AB Since AC is perpendicular to AB, the gradient of AC is negative reciprocal of, i.e. Gradient of AC - - Hence equation of AC: y 6 - ( 4) y () y --- () Since both lines intersect, sub () into (): (- + 4) (Ans) Sub 6 into (): y - (6) Hence, C (6, ) [6] [N97/I/] The points A and B have coordinates (, ) and (0, 8) respectively. Find the equation of the perpendicular bisector of AB. Midpoint of AB y + 0, 0, + 8 [4]

6 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS y (6, 5) Gradient of bisector - Equation of bisector: Gradient of AB Y ( 6) 3 3y 5-4( 6) y (Ans) 3 4 [N000/I/] A (5, ) P C B (0, ) O The line joining A(5, ) and B(0, ) meets the -ais at C. The point P lies on AC such that AP:PB 3:. (i) Find the coordinates of P By similar triangles, we drop lines from A and P to form the following figure: Q A [5] P E B F C O Q Let P be (, y). By similar triangles,

7 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS By similar triangles, PE BF AP AB AP AP + PB AE AF -y - - y 0 - y 0 AP AB y 6 y 5 Hence, P (, 5) A line through P meets the -ais at Q and angle PCQ angle PQC. Find: (ii) The equation of PQ Since angle PCQ angle PQC, then gradient of AC - gradient of PQ Gradient of PQ - Gradient of AC Hence, equation of PQ: y ( ) y y (Ans) (ii) The coordinates of Q Since Q lies on -ais, then y 0. Thus, let Q (Q 0) and sub into equation of PQ: 0 -(Q ) (Q ) Q 9 Q 4.5 Hence, Q (4.5, 0) [N00/I/3]

8 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS y 3 y 3-5 B A (, 6) y ½ C D O The diagram, which is not drawn to scale, shows a parallelogram OABC where O is the origin and A is the point (, 6). The equations of OA, OC and CB are y 3, y ½ and y 3 5 respectively. The perpendicular from A to OC meets OC at the point D. Find (i) The coordinates of C, B and D. [8] To find coordinates of C, equate y ½ and y 3 5: ½ ½.5 6 (Ans) Sub 6 into y ½: y ½ (6) 3 (Ans) Hence, C (6, 3) Since AB is parallel to y ½, then gradient of AB is also ½. Equation of AB: y 6 ½ ( ) ½ y ½ + 6 y ½ + 5 (Ans) Since B is intersection point of y ½ + 5 and y 3 5, equate both to get B: ½ (Ans) Sub 8 into y ½ + 5: y ½(8) (Ans) Hence B (8, 9) Equation of AD: Gradient of AD Gradient of OC

9 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS y ( ) y y y To find coordinates of D, equate y and y ½: ½ (Ans) Sub 4 into y ½: y ½(4) (Ans) Hence D (4, ) (ii) The perimeter of the parallelogram OABC, correct [3] to decimal place Length of OA Length of OC Perimeter of OABC (Length of OA + Length of OC) ( ) units 6. units (Ans) [N003/I/] A parallelogram ABCD passes through vertices A(-3, ), B(4, 9) and C(, -3). Find: (a) The midpoint of the diagonal AC Midpoint of AC , + (-3), - (4, -) (Ans) (b) The fourth verte D Since ABCD is a parallelogram, the midpoint of its diagonals would bisect each other (i.e. they would both be the same). Hence, let D be (h, k). Midpoint of AC Midpoint of BD (4, -) Focusing on coordinate only: 4 + h 4 + h, 9 + k Focusing on y coordinate only: h k h 8 4 k Hence, D (4, ) (Ans) 9 + k

10 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS G Graphs of Equations y a n and y k For such questions, you would only be required to: - Draw the graphs given the values - Find the solution set to a particular equation by adding a suitable straight line - Find the corresponding or y coordinate, given one. As such, refer to your notes from Mathematics on Functions and Graphs for how to answer these questions. For reference purposes, here are the various y a n and y k graphs. y a - y a y a - y a y a y a 3 y a y a y a 3 3 y a y a - y a y a - 3 y a 3 y k (where k > 0) y k (where k < 0)

11 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS H Equation of a Circle We know that the length of a line is - + y - y. We now have a circle with a radius CF of length r, where C (a, b) is the centre of the circle and F (, y) is a point on the circumference of the circle. - a + y - b r Taking square on both sides, ( a) + (y b) r We thus get the equation that a circle of centre C (a, b) would have an equation of ( a) + (y b) r where r is the radius. Now consider a circle with the equation: + y + g + fy + c 0. How can we determine the centre and the radius? Let us try manipulating the equation. + y + g + fy + c 0 + g + y + fy - c + (g)() + y + (f)(y) - c Complete the square: + (g)() + g g + y + (f)(y) + f f - c ( + g) g + (y + f) f - c ( + g) + (y + f) - c + g + f ( + g) + (y + f) - c + g + f We thus gather that the centre of the circle is (-g, -f) and the radius is g + f - c. In summary, FORM Property FORM (a, b) Centre of a Circle, C (-g, -f) r Radius of a Circle, r g + f - c ( a) + (y b) r Equation of a Circle + y + g + fy + c 0 WORKED EXAMPLES The equation of a circle, C, is + y 6 8y (i) Find the coordinates of the centre of C and find the radius of C. [3] + y 6 8y y 8y - 6 (3)() + y ()(4)(y) - 6 (3)() y ()(4)(y) ( 3) 3 + (y 4) 4-6 ( 3) 9 + (y 4) 6-6 ( 3) + (y 4) 5-6 ( 3) + (y 4) Hence, coordinates of centre of C is (3, 4) and radius of C is 3 units. (ii) Show that C touches the y-ais [] When C touches y-ais, 0. Hence, sub 0 into equation. (0 3) + (y 4) (y 4) 9 (y 4) 0 y y 4

12 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS Therefore, as there is a corresponding y coordinate for the value of 0, C does touch the y-ais. (iii) Find the equation of the circle which is a reflection of C in the y-ais. [3] Centre of C (3, 4). Hence, a circle which has a reflection of C in the y-ais has a centre of (-3, 4). The radius is the same as it the same size as C. ( (-3)) + (y 4) 3 ( + 3) + (y 4) 9 X y 8y X y 8y X + y + 6 8y (Ans) [SP008/I/7] C y O C A The diagram shows two circles, C and C. Circle C has its centre at the origin O. Circle C passes through O and has its centre at Q. The point P (8, -6) lies on both circles and OP is a diameter of C. Q P (8, -6) (i) Find the equation of C. [] Since centre of circle is origin, it is (0, 0) Radius of circle C Diameter of Circle C Length of OP (0 + 8) + (0-6) 8 + (-6) units Hence, equation of C : ( 0) + (y 0) 0 + y 00 + y 00 0 (Ans) (ii) Find the equation of C. [3] Centre of C Midpoint of OP, Q 0 + 8, 0 + (-6) 8 (4, -3) Radius of C ½ (Length of OP) ½ (0 units) 5 units Hence, equation of C : ( 4) + (y (-3)) 5 ( 4) + (y + 3) y + 6y y 8 + 6y y 8 + 6y 0 (Ans) (iii) B The line through Q perpendicular to OP meets the circle C at the points A and B. Show that the coordinates of A and B are a + b 3 [7]

13 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS and a - b 3 respectively, where a and b are integers to be found. + y () Gradient of AB - Equation of AB: y (-3) Gradient of OP ( - 4) y y Sub into : ( (4)(5) ± (-8) - 4()(-) () 8 ± ± 08 4 ± 7 4 ± or (proven) [N008/II/]

14 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS I Linear Law (Source: New Syllabus Additional Mathematics 8 th Edition (007), Dr Yeap Ban Har/ Teh Keng Seng/ Loh Cheng Yee) In research work, when two variables are believed to be related, a set of corresponding values are obtained. Any equation must be in the form of Y mx + c, whereby: - Y is a function of y (i.e. must contain y in any form such as ln y, sin y, y, ) y - m is a constant term which is the gradient (i.e. cannot contain or y terms) - X is a function of (i.e. must contain y in any form such as lg, cos, y, y ) - c is a constant term which is the Y-intercept (i.e. cannot contain or y terms) However, not all eperimental results obey a linear relationship. Eamples of such graphs include: y a b y a + b y a + b Where a and b are constants With a little algebraic manipulation, each of the above relationships can be converted to a linear relationship. The table below shows the manipulation of some common graph types. y a + b y a + b Divide throughout by y a + b Hence, a straight line graph is obtained when y (Y) is plotted against (X). Here, a denotes the Y-intercept and b denotes the gradient. y a + b y a + b y a( ) + b Hence, a straight line graph is obtained when y (Y) is plotted against (X). Here, a denotes the gradient and b denotes the Y-intercept. y a + b y a + Multiply throughout by y a + b Hence, a straight line graph is obtained when y (Y) is plotted against (X). Here, a denotes the gradient and b denotes the Y-intercept. y a + b y a + b a Divide throughout by y + b y a( ) + b Hence, a straight line graph is obtained when y (Y) is plotted against (X). Here, a denotes the gradient and b denotes the Y-intercept. b

15 y 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS y ae b y ae b Taking ln on both sides ln y ln (ae b ) Using product law to split up ln (ae k ) ln y ln a + ln e b Using power law to change ln e k ln y ln a + b ln e Since ln e ln y ln a + b Hence, a straight line graph is obtained when ln y (Y) is plotted against (X). Here, ln a denotes the Y-intercept and b denotes the gradient. y a b y a b Taking lg on both sides lg y lg (a b ) Using product law to split up lg (a b ) lg y lg a + lg b Using power law to change lg b lg y lg a + b lg Hence, a straight line graph is obtained when lg y (Y) is plotted against lg (X). Here, lg a denotes the Y-intercept and b denotes the gradient. y a y a Divide throughout by y y a( ) a Hence, a straight line graph is obtained when y (Y) is plotted against (X). Here, a denotes the gradient and 0 denotes the Y-intercept (i.e. passes through origin O). WORKED EXAMPLE (a) The table shows eperimental values of two variables, and y y Using the vertical ais for y and the horizontal ais for, plot y against and obtain a straight line graph. Use your graph to: [0] Plot a new table of values: y Using new values, plot the graph and draw a line of best fit y (i) Epress y in terms of To find gradient, choose two points on the graph (e.g. (.0, 7.0) and (5.0, 55.0)

4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS Gradient 55.0 7.0 5.0.0 48.0 4.0 From the graph, Y-intercept 5.

16 4038 ADDITIONAL MATHEMATICS TOPIC : GEOMETRY AND TRIGONOMETRY SUB-TOPIC.: COORDINATE GEOMETRY IN TWO DIMENSIONS Gradient From the graph, Y-intercept 5.0 Hence, Y mx + c y ( ) y + 5 (Ans) 30 y y 30 Hence, from the graph, when y 30, (Ans to 3sf) Variables and y are related in such a way that, when y is plotted against, a straight line is produced which passes through the points A(4, 6), B(3, 4) and P(p, 4.48), as shown in the diagram. Find [6] (i) y in terms of To find gradient, choose the two given points on the graph (i.e. (4, 6) and (3, 4) Gradient Hence, equation of line: Y Y m(x X ) (y ) (4) ( 3) y 4 6 y (Ans) (ii) The value of p Find value of p when (y ) 4.48 y + y p 3.4 (Ans) (iii) The value of and of y at the point P. At P, (Ans) At P, y 4.48 Y y (Ans) [N000/II/4]

Education Resources. This section is designed to provide examples which develop routine skills necessary for completion of this section.

Education Resources. This section is designed to provide examples which develop routine skills necessary for completion of this section. Education Resources Straight Line Higher Mathematics Supplementary Resources Section A This section is designed to provide examples which develop routine skills necessary for completion of this section.