6. COORDINATE GEOMETRY

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1 6. CRDINATE GEMETRY Unit 6. : To Find the distance between two points A(, ) and B(, ) : AB = Eg. Given two points A(,3) and B(4,7) ( ) ( ). [BACK T BASICS] E. P(4,5) and Q(3,) Distance of AB = (4 ) (7 3) PQ = = 4 6 = 0 unit. E. R(5,0) and S(5,) E3. T(7,) and U(,5) [ 0 ] E4. V(0,6) and W(4,) [] E5. X(-4,-) and Y(-,) [ 4 ] [ 5 ] [ 8 ] More challenging Questions. E. The distance between two points A(, 3) and B(4, k) is 5. Find the possible vales of k. E. The distance between two points P(-, 3) and Q(k, 9) is 0. Find the possible values of k. 7, - E3. The distance between two points R(-, 5) and S(, k) is 0. Find the possible vales of k. 7, -9 E4. The distance between two points K(-, p) and L(p, 9) is 50. Find p. 6, 4 E5. The distance between two points U(4, -5) and V(, t)is 0. Find the possible vales of t. p=0,6 E6. If the distance between (0, 0) and P(k, k) is the same as the distance between the points A(-4, 3) and B(, -7), find the possible values of k. t=-9, - k= 5,-5 6. Coordinate Geometr

2 Unit 6. : Division of a Line Segment 6.. To find the mid-point of Two Given Points. Formula : Midpoint M =, Eg. P(3, ) and Q(5, 7) E P(-4, 6) and Q(8, 0) Midpoint, M= 3 5 7, = (4, 9 ) E P(6, 3) and Q(, -) E3 P(0,-), and Q(-, -5) (, 3) (4, ) (- ½,-3) 6.. Division of a Line Segment Q divides the line segment PR in the ratio PQ : QR = m : n. P(, ),R(, ) m P(, ) Q(, ) R(, ) n m n m Q (,) =, m n m n n P(, ) m n Q(, ) R(, ) (NTE : Students are strongl advised to sketch a line segment before appling the formula) Eg. The point P internall divides the line segment joining the point M(3,7) and N(6,) in the ratio :. Find the coordinates of point P. E. The point P internall divides the line segment joining the point M (4,5) and N(-8,-5) in the ratio : 3. Find the coordinates of point P. P(, ) N(6, ) M(3, 7) (3) (6) (7) () P=, 5 =, 3 3 = 5, 3, 5 6. Coordinate Geometr

3 More Eercise : The Ratio Theorem (NTE : Students are strongl advised to sketch a line segment before appling the formula) E. R divides PQ in the ratio :. Find the coordinates of R if E. P divides AB in the ratio 3 :. Find the coordinates of P if (a) P(, ) and Q( -5, ) (c) A(, -3) and B( -8, 7) (b) P(-4, 7) and Q(8, -5) (d) A(-7, 5) and B(8, -5) (a) (-3, 8) (b) (4, -) E3. M is a point that lies on the straight line RS such that 3RM = MS. If the coordinates of the points R and S are (4,5) and (-8,-5) respectivel, find the coordinates of point M. (a) (-4, -3) (b) (, -) E4. P is a point that lies on the straight line TU such that 3TP = PU. If the coordinates of the points T and U are (-9,7) and (,-3) respectivel, find the coordinates of point P. 3RM = MS RM =, RM: MS= : 3 MS 3 Ans : 5, E5. The points P(3, p), B(-, ) and C(9,7) lie on a straight line. If P divides BC internall in the ratio m : n, find (a) m : n, (b) the value of p. (-5, 3) E6. R(, ), divides the points P(k, k) and Q(, 4) in the ratio 3 : 5. Epress in terms of. (a) : 3 (b) p = 4 (Ans : = 4) 6. Coordinate Geometr 3

4 Unit 6.3 To Find Areas of Polgons Area of a polgon = Note : The area found will be positive if the coordinates of the points are written in the anti-clockwise order, and negative if the are written in the clock-wise order. Eample : Calculate the area of a triangle given : E. P(0, ), Q(, 3) and R(,5). P(,3), Q(5,6) and R(-4,4) Area of PQR = = Area of PQR = = unit. The coordinates of the triangle ABC are (5, 0), (,) and (8, k) respectivel. Find the possible values of k, given that the area of triangle ABC is 4 units. 7 unit 3. The coordinates of the triangle RST are (4, 3), (-, ) and (t, -3) respectivel. Find the possible values of t, given that the area of triangle RST is units. k=3,35 t=0,- ii) Area of a quadrilateral = P(,5), Q(4,7), R(6,6) and S(3,).. P(,-), Q(3,3), R(-, 5) and S(-4, -). Area of PQRS = = 8 unit [7] Note : If the area is zero, then the points are collinear.. Given that the points P(5, 7), Q(4, 3) and R(-5, k) are collinear, find the value of k.. Show that the points K(4, 8), L(, ) and M(, -) are collinear. k= Coordinate Geometr 4

5 Unit 6.4 : Equations of Straight Lines The Equation of a Straight line ma be epressed in the following forms: i) The general form : a + b + c = 0 ii) The gradient form : =m+c; m = gradient, c = -intercept iii) The intercept form : + =, a b a = -intercept, b = -intercept a) If given the gradient and one point: = m ) ( Gradient = m P(, ) E. Find the equation of a straight line that passes through the point (5,) and has a gradient of -. Eg. Find the equation of a straight line that passes through the point (,-3) and has a gradient of 4. = m( ) ( 3) ( ) E. Find the equation of a straight line that passes through the point (-8,3) and has a gradient of 4 3. =- + b) If two points are given : Note : You ma find the gradient first, then use either (a) =m+c r (b) =m( ) r (c) = 4 = Eg. Find the equation of a straight line that passes through the points (-3, -4) and (-5,6) ( 4) = ( 3) 6 ( 4) 5 ( 3) E. Find the equation of a straight line that passes through the points (, -) and (3,0) 5 9 E. Find the equation of a straight line that passes through the points (-4,3) and (,-5) = = 0 6. Coordinate Geometr 5

6 c) The-intercept and the -intercept are given: int ercept m = - int errcept Equation of Straight Line is : + = a b Note : Sketch a diagram to help ou! At the -ais, =0 At the -ais, =0 E. The -intercept and the -intercept of the straight line PQ are -6 and 3 respectivel. Find the gradient and the equation of PQ. E. The -intercept and the -intercept of the straight line PQ are 4 and -8 respectivel. Find the gradient and the equation of PQ. int ercept m PQ = int errcept = 8 4 = 4 Equation : + = E3. The -intercept of a straight line AB is -5 and its gradient is -3. Find the -intercept of the straight line AB and the equation of AB. = +6 Etra Vitamins for U. Find the gradient and the equation of AB. A - 6 B = 0. The -intercept of a straight line RS is and its gradient is 3. Find the -intercept of the straight line RS and the equation ofrs. 3 = 6 3. Find the equation of KL in the intercept form. K 3 = Find the equation of the line which connects the origin and the point S (-, 6). 6 L For Q3 above, write down the equation of KL in the general form. = 3 6. Write down the equation of the straight line which passes through the points P(3, ) and Q (3, 8). + 6=0 [ = 3] 6. Coordinate Geometr 6

7 Unit 6.5 Parallel Lines and Perpendicular lines 6.5. Parallel lines, m = m 6.5. Perpendicular lines, m m = - Unit 6.5. Determine whether each of the following pairs of lines are parallel. Eg. = 3 and 3 =4. = +5 and 4 + = 5 =3, m = 3 3 =4 = 3 4, m = 3 Since m =m, the two line are parallel = 7 and = = 5 and 6 = N 4. 3 = and 6 = 3 + N 5. 4 and 8 = Y Y N Unit 6.5. Determine whether each of the following pairs of lines are perpendicular. Eg. = 3 and +3=4. = +5 and 4 + = 9 =3, m = 3 +3= 4 3 = +4 4, m = Since m.m = 3, 3 The two given lines are perpendicular. N. 3 = and + 3 = 3. 3 = and 6 + = = - 3 and N 5. and =0 3 4 Y Y Y 6. Coordinate Geometr 7

8 6.5. Applications (m.m = ) E. (SPM 004). Diagram shows a straight line PQ with the equation. Find the 4 equation of the straight line perpendicular to PQ and passing through the point Q. E.. Diagram shows a straight line PQ with the equation. Find the equation of the 6 straight line perpendicular to PQ and passing through the point P. Q Diagram Q Diagram P P Answer: Answer: =½+4 E.3 Diagram 3 shows a straight line RS with the equation + = 6. Find the equation of the straight line perpendicular to RS and passing through the point S. =3 8 E.4. Diagram 4 shows a straight line AB with the equation 3 = 6. Find the equation of the straight line perpendicular to AB and passing through the point B. R Diagram 3 B S A Diagram 4 Answer: Answer: = = Coordinate Geometr 8

9 6.5. Applications (m.m = ) more eercises E.5 Diagram 5 shows a straight line PQ with the equation =. Find the equation of the straight line perpendicular to RS and passing through the midpoint of RS. R Diagram 5 E.6. Diagram 6 shows a straight line AB with the equation. Find the equation of the 4 6 perpendicular bisector of the line AB. B S A Diagram 6 Answer: Answer: 4+3 = 8 E.7. Find the equation of the straight line that passes through the point (, ) and is perpendicular to the straight line = = 6 E.8 Find the equation of the straight line that passes through the point (3, 0) and is perpendicular to the straight line 3 =. =3 E.9 Find the equation of the straight line that passes through the origin and is perpendicular to the straight line that passes through the points P(, )andq(-3,7). +3 = 6 E. 0 Find the equation of the straight line that passes through the point (-,4) and is perpendicular to the straight line which passes through the origin and the point (6, ). =½ =-3 6. Coordinate Geometr 9

10 Unit 6.6 Equation of a Locus Note : Students MUST be able to find distance between two points [ using Pthagoras Theorem] TASK : To Find the equation of the locus of the moving point P such that its distances of P from the points Q and R are equal. Eg. Q(6, -5) and R(,9) Let P = (,), then PQ = PR ( 6) ( ( 5) = ( ) ( 9) Square both sides to eliminate the square roots. 6) ( 5 = ) ( Q(6, -5) P(, ) Locus of P R(, 9) E. Q(,5) and R(4,) E. Q(-3, 0) and R(6, 4) E3. Q(, -3) and R(-4, 5) =0 E4. Q(6, -) and R(0, ) = = 0 More challenges. E5. Given two points A(3, ) and B(7, -4). Find the equation of the perpendicular bisector of AB. 3 9=0 E6. Given two points P(4, 0) and QB(-6, 0). Find the equation of the the perpendicular bisector of PQ. 3 = - 3 +=4 6. Coordinate Geometr 0

11 TASK : To find the equation of the locus of the moving point P such that its distances from the points A and B are in the ratio m : n (Note : Sketch a diagram to help ou using the distance formula correctl) Eg. A(-,3), B(4,8) and m : n = : Let P=(, ) LP PM = LK = KM ( ( )) ( 3) = ( ) ( 3) 4 ( 4) ( = 4) ( 8 4( ) ( 3 ) = 4 ( 8) ) is the equation of locus of P. A(-, 3) 6 64 P(, ) B(4, 8) E. A(, 5), B(4, ) and m : n = : E. A(-3, ), B(3, ) and m : n = : = 0 E3. A(, 3), B(-, 6) and m : n = : =0 E4. A(5, -), B(-4, ) and m : n = : =0 E5. P(-, 3), Q( 4, -) and m : n = : =0 E6. A(, 5), B(-4, -5) and m : n = 3 : = = 0 6. Coordinate Geometr

12 SPM FRMAT QUESTINS. (003) The equations of two straight lines are and 5 = Determine whether the lines are perpendicular to each other.. The equations of two straight lines are 4 and 3 = + 6. Determine 3 whether the lines are perpendicular to each other. [Y] 3.(004) Diagram 4 shows a straight line PQ with the equation. Find the equation of the 3 straight line perpendicular to PQ and passing through the point Q. [N] 4. Diagram 5 shows a straight line RS with the equation. Find the equation of the 6 4 straight line perpendicular to RS and passing through the point S. Q Diagram 4 R Diagram 5 P S [ 3 ] 3 5. (005) The following information refers to the equations of two straight lines, JK and RT, which are perpendicular to each other. JK : = p + k RT : = (k ) + p where p and k are constants. Epress p in terms of k. [ = 3-8] 6. The following information refers to the equations of two straight lines, PQ and RS, which are perpendicular to each other. PQ : p += k RS : = (k ) + p where p and k are constants. Epress p in terms of k. p k p k 6. Coordinate Geometr

13 7. (006) Diagram 5 shows the straight line AB which is perpendicular to the straight line CB at the point B. 8.Diagram 6 shows the straight line PQ which is perpendicular to the straight line RQ at the point Q. A(0, 4) B Diagram 5 P(0, 6) Q Diagram 6 C R The equation of CB is =. Find the coordinates of B. The equation of QR is =4. Find the coordinates of Q (, 3) 9.(004) The point A is (-, ) and B is (4, 6). The point P moves such that PA : PB = : 3. Find the equation of locus of P. [3 marks] Q(5, ) 0. The point R is (3, -5) and S is (0, ). The point P moves such that PR : PS = :. Find the equation of locus of P. [3 marks] [ =0].The point A is (8, -) and B is (4, 6). Find the equation of the perpendicular bisector of AB. [3 marks] [ = 0]. The point R is (, -3) and S is (4, 5). The point P moves such that it is alwas the same distance from R and from S. Find the equation of locus of P. [3 marks] = +4 = 7 6. Coordinate Geometr 3

1 k. cos tan? Higher Maths Non Calculator Practice Practice Paper A. 1. A sequence is defined by the recurrence relation u 2u 1, u 3.

1 k. cos tan? Higher Maths Non Calculator Practice Practice Paper A. 1. A sequence is defined by the recurrence relation u 2u 1, u 3. Higher Maths Non Calculator Practice Practice Paper A. A sequence is defined b the recurrence relation u u, u. n n What is the value of u?. The line with equation k 9 is parallel to the line with gradient