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1 9 CARTESIAN SYSTEM OF COORDINATES You must have searched for our seat in a cinema hall, a stadium, or a train. For eample, seat H-4 means the fourth seat in the H th row. In other words, H and 4 are the coordinates of our seat. Thus, the geometrical concept of location is represented b numbers and alphabets (an algebraic concept). Also a road map gives us the location of various houses (again numbered in a particular sequence), roads and parks in a colon, thus representing algebraic concepts b geometrical figures like straight lines, circles and polgons. The stud of that branch of Mathematics which deals with the interrelationship between geometrical and algebraic concepts is called or Cartesian in honour of the famous French mathematician Rene Descartes. In this lesson we shall stud the basics of coordinate geometr and relationship between concept of straight line in geometr and its algebraic representation. OBJECTIVES After studing this lesson, ou will be able to: define Cartesian Sstem of s including the origin, coordinate aes, quadrants, etc; derive distance formula and section formula; derive the formula for area of a triangle with given vertices; verif the collinearit of three given points; state the meaning of the terms : inclination and slope of a line; MATHEMATICS 07 Get Discount Coupons for our Coaching institute and FREE Stud Material at

2 find the formula for the slope of a line through two given points; state the condition for parallelism and perpendicularit of lines with given slopes; find the intercepts made b a line on coordinate aes; define locus as the path of a point moving in a plane under certain conditions; and find the equation of locus under given conditions. EXPECTED BACKGROUND KNOWLEDGE Number sstem. Plotting of points in a coordinate plane. Drawing graphs of linear equations. Solving sstems of linear equations. 9. RECTANGULAR COORDINATE AXES Recall that in previous classes, ou have learnt to fi the position of a point in a plane b drawing two mutuall perpendicular lines. The fied point O,where these lines intersect each other is called theorigin O as shown in Fig. 9. These mutuall perpencular lines are called the coordinate aes. The horizontal line XOX' is the-ais or ais of and the vertical line YOY' is the - ais or ais of. 9.. CARTESIAN COORDINATES OF A POINT To find the coordinates of a point we proceed as follows. Take X'OX and YOY' as coordinate aes. Let P be an point in this plane. From point P draw ' PA XOX and PB YOY '. Then the distance OA measured along -ais and the distance OB measured along -ais determine the position of the point P with reference to these aes. The distance OA measured along the ais of is called the abscissa or -coordinate and the distance OB (PA) measured along ais is called the ordinate or -coordinate of the point P. The abscissa and the ordinate taken together are called the coordinates of the point P. Thus, the coordinates of the point P are ( and ) which represent the position of the point P point in a plane. These two numbers are to form an ordered pair beacuse the order in which we write these numbers is important. ' ' O ' Fig. 9. B ' O A P (, ) Fig MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

3 In Fig. 9. ou ma note that the position of the ordered pair (,) is different from that of (,). Thus, we can sa that (,) and (,) are two different ordered pairs representing two different points in a plane. O (,) (,) ' ' Fig QUARDRANTS We know that coordinate aes XOX' and YOY' divide the region of the plane into four regions. These regions are called the quardrants as shown in Fig In accordance with the convention of signs, for a point P (,) in different quadrants, we have I quadrant : > 0, > 0 II quadrant : < 0, > 0 III quadrant : < 0, < 0 IV quadrant : > 0, < 0 9. DISTANCE BETWEEN TWO POINTS Recall that ou have derived the distance formula between two points P (, ) and Q (, ) in the following manner: Let us draw a line l XX' through P. Let R be the point of intersection of the perpendicular from Q to the line l. Then PQR is a right-angled triangle. Also PR M M OM OM and QR QM RM ' N N ' O(0,0) ' II Quadrant III Quadrant M ' Fig. 9.4 P (, ) I Quadrant IV Quadrant M R Fig. 9.5 Q(, ) MATHEMATICS 09 Get Discount Coupons for our Coaching institute and FREE Stud Material at

4 QM PM ON ON Now PQ PR + QR (Pthagoras theorem) ( ) + ( ) PQ ( ) + ( ) Note : This formula holds for points in all quadrants. Also the distance of a point P(,) from the origin O(0,0) is OP +. Let us illustrate the use of these formulae with some eamples. Eample 9. Find the distance between the following pairs of points : (i) A(4,) and B(0,6) (ii) M(,) and N(0, 6) Solution : (i) Distance between two points ( ) ( ) Here 4,, 0, Distance between A and B ( ) ( ) 5 ( 4 ) + ( ) Distance between A and B is 5 units. (ii) Here,, 0 and 6 Distance between A and B ( ) ( ) ( ) + ( 8) 0 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

5 Distance between M and N units Eample 9. Show that the points P(, ), Q(, ) and R (, 6) are the vertices of a right-angled triangle. Solution: PQ ( + ) + ( + ) QR ( 4) + () and RP + ( 7) PQ + QR RP PQR is a right-angled triangle (b converse of Pthagoras Theorem) Eample 9. Show that the points A(, ), B(4, 5) and C(, 0) lie on a straight line. Solution: Here, AB ( ) ( ) + 5 units 4 8 units units BC ( ) ( ) 50 units 5 units units and AC ( ) ( ) + 0 units units units MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

6 Now AB + AC ( ) + units 5 units BC i.e. BA + AC BC Hence, A, B, C lie on a straight line. In other words, A,B,C are collinear. Eample 9.4 Prove that the points (a, 4a), (a, 6a) and ( a a,5a) an equilateral triangle whose side is a. + are the vertices of Solution: Let the points be A (a, 4a), B (a, 6a) and C ( a + a,5a) AB 0 + (a) a units BC ( a) + ( a) units a + a a units and AC ( a) + ( + a) a units AB + BC > AC, BC + AC > AB and AB + AC > BC and AB BC AC a A, B, C form the vertices of an equilateral triangle of side a. CHECK YOUR PROGRESS 9.. Find the distance between the following pairs of points. (a) (5, 4) and (, ) (b) (a, a) and (b, b). Prove that each of the following sets of points are the vertices of a right angled-trangle. (a) (4, 4), (, 5), (, ) (b) (, ), (0, ), (, ). Show that the following sets of points form the vertices of a triangle: (a) (, ), (, ) and (0, 0) (b) (0, a), (a, b) and (0, 0) (if ab 0) 4. Show that the following sets of points are collinear : (a) (, 6), (, 4) and ( 4, 8) (b) (0,), (0, 4) and (0, 6) 5. (a) Show that the points (0, ), (, ), (6, 7) and (8, ) are the vertices of a rectangle. (b) Show that the points (, ), (6, ), (, 4) and (0, ) are the vertices of a square. MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

7 9. SECTION FORMULA 9.. INTERNAL DIVISION Let P, ) and Q, ) be two given points on a line l and R (, ) divide PQ ( ( internall in the ratio m : m. To find : The coordinates and of point R. Construction : Draw PL, QN and RM perpendiculars to XX' from P, Q and R respectivel and L, M and N lie on XX'. Also draw RT QN and PV QN. Method : R divides PQ internall in the ratio m : m. R lies on PQ and PR RQ m m Also, in triangles, RPS and QRT, RPS QRT (Corresponding angles as PS RT ) 0 and RSP QTR 90 RPS ~ QRT (AAA similarit) PR RS PS... (i) RQ QT RT Also, PS LM OM OL RT RS QT MN ON OM RM SM QN TN. From (i), we have m m m ( ) m ( ) and m ) m ( ) ( P (, ) ' O O(0,0) ' m R (, ) m L Q (, ) S M N T V Fig. 9.6 m + m m + m and m + m m + m MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

8 Thus, the coordinates of R are: m + m m + m, m + m m + m s of the mid-point of a line segment If R is the mid point of PQ, then, m m (as R divides PQ in the ratio : + + s of the mid point are, 9.. EXTERNAL DIVISION Let R divide PQ eternall in the ratio m :m To find : The coordinates of R. Construction : Draw PL, QN and RM perpendiculars to XX' from P, Q and R respectivel and PS RM and QT RM. Clearl, RPS ~ RQT. RP PS RS RQ QT RT or m m m ( ) m ( ) and m ) m ( ) These give: ( m m m m m m and m m Hence, the coordinates of the point of eternal division are P(, ) ' O Q (, ) ' m L S R (, ) N M T S Fig. 9.7 m m m m, m m m m Let us now take some eamples. 4 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

9 Eample 9.5 Find the coordinates of the point which divides the line segment joining the points (4, ) and (, 5) internall and eternall in the ratio :. Solution: (i) Let P (, ) be the point of internal division. ( ) + (4) 6 (5) + ( ) 4 and P has coordinates, 5 5 If Q (', ') is the point of eternal division, then ()( ) (4) ' 8 and ()(5) ( ) ' 6 Thus, the coordinates of the point of eternal division are (8, 6). Eample 9.6 In what ratio does the point (, ) divide the line segment joining the points (,4) and (, 6)? Solution : Let the point P(, ) divide the line segement in the ratio k :. k + 6k + 4 Then the coordinates of P are, k + k + But the given coordinates of P are (, ) k + k + k + k + k P divides the line segement eternall in the ratio :. Eample 9.7 The vertices of a quadrilateral ABCD are respectivel (, 4), (,), (0, ) and (, ). If E, F, G, H are respectivel the midpoints of AB, BC, CD and DA, prove that the quadrilateral EFGH is a parallelogram. Solution : Since E, F, G, and H, are the midpoints of the sides AB, BC, CD and DA, therefore, the coordinates of E, F, G, and H respectivel are : 4 +,, ,,, and + 4 +, MATHEMATICS 5 Get Discount Coupons for our Coaching institute and FREE Stud Material at

10 5 E,, F(,0), G, and H (, ) are the required points. Also, the mid point of diagonal EG has coordinates + 5, +, s of midpoint of FH are + 0 +,, Since, the midpoints of the diagonals are the same, therefore, the diagonals bisect each other. Hence EFGH is a parallelogram. CHECK YOUR PROGRESS 9.. Find the midpoint of each of the line segements whose end points are given below: (a) (, ) and (, 5) (b) (6,0) and (,0). Find the coordinates of the point dividing the line segment joining ( 5, ) and (, 6) internall in the ratio :.. (a) Three vertices of a parallelogram are (0,), (0,6) and (,9). Find the fourth verte. (b) (4, 0), ( 4, 0), (0, 4) and (0, 4) are the vertices of a square. Show that the quadrilateral formed b joining the midpoints of the sides is also a square. 4. The line segement joining (, ) and (5, ) is trisected. Find the points of trisection. 5. Show that the figure formed b joining the midpoints of the sides of a rectangle is a rhombus. 9.4 AREA OF A TRIANGLE Let us find the area of a triangle whose A(, ) vertices are A(, ), B(, ) and C(, ) C (, ) Draw AL, BM and CN perpendiculars to XX'. B(, ) area of ABC ' O M L N ' Fig MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

11 Area of trapzium. BMLA + Area of trapzium. ALNC Area of trapzium. BMNC ( BM + AL) ML + ( AL + CN ) LN ( BM + CN ) MN ( + )( ) + ( + )( ) ( + )( ) [ ] ) + ( ) + ( ) ( [ ( ) + ( ) + ( )] This can be stated in the determinant form as follows : Area of ABC Eample 9.8 Find the area of the triangle whose vertices are A(, 4), B(6, ) and C( 4, 5). Solution: The area of 4 ABC [ ( + 5) 4(6 + 4) + ( 0 8) ] 69 [ ] As the area is to be positive 69 Area of ABC square units Eample 9.9 If the vertices of a triangle are (, k), (4, ) and ( 9, 7) and its area is 5 square units, find the value(s) of k. Solution : Area of triangle k MATHEMATICS 7 Get Discount Coupons for our Coaching institute and FREE Stud Material at

12 [ 7 k(4 + 9) + (8 7) ] [ 0 k + ] [ 9 k] Since the area of the triangle is given to be5, 9 k 5 or, 9 k 0 k 9 or, k CHECK YOUR PROGRESS 9.. Find the area of each of the following triangles whose vertices are given below : () (0, 5), (5, 5), and (0,0) (b) (, ), (, ) and (, ) (c) (a, 0), (0, a) and (0, 0). The area of a triangle ABC, whose vertices are A (, ), B(, ) and C 5, k is sq unit. Find the value of k. Find the area of a rectangle whose vertices are (5, 4), (5, 4), ( 5, 4) and ( 5, 4) 4. Find the area of a quadrilateral whose vertices are (5, ), (4, 7), (, ) and (, 4) 9.5 CONDITION FOR COLLINEARITY OF THREE POINTS The three points A, ), B(, ) and C(, ) are collinear if and onl if the area of the triangle ABC becomes zero. ( i.e. [ + + ] i.e In short, we can write this result as 0 8 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

13 0 Let us illustrate this with the help of eamples: Eample 9.0 Show that the points A( a, b + c), B( b, c + a) and C( c, a + b) are collinear. Solution : Area of triangle ABC Hence the points are collinear. a b + c b c + a c a + b a + b + c a + b + c a + b + c b + c c + a a + b b + c ( a + b + c ) c + a 0 a + b Eample 9. For what value of k, are the points (, 5), (k, ) and (4, ) collinear? Solution : Area of the triangle formed b the given points is k 4 5 k [ 0 5 k ] Since the given points are collinear, therefore [ 6 k + 6 ] k + k + 0 k Hence, for k, the given points are collinear. MATHEMATICS 9 Get Discount Coupons for our Coaching institute and FREE Stud Material at

14 CHECK YOUR PROGRESS 9.4. Show that the points (, ), (5, 7) and (8, ) are collinear.. Show that the points (, ), (5, ) and (6, 4) are collinear.. Prove that the points (a, 0), (0, b) and (, ) are collinear if +. a b 4. If the points ( a, b),( a, b ) and ( a a, b b ) are collinear, show that a b ab 5. Find the value of k for which the points (5, 7), (k, 5) and (0, ) are collinear. 6. Find the values of k for which the point (k, k), ( k+, k) and ( 4 k, 6 k) are collinear. 9.6 INCLINATION AND SLOPE OF A LINE Look at the Fig The line AB makes an angle or anticlockwise direction). π + α with the -ais (measured in The inclination of the given line is represented b the measure of angle made b the line with the positive direction of -ais (measured in anticlockwise direction) In a special case when the line is parallel to-ais or it coincides with the-ais, the inclination of the line is defined to be 0 0. (a) Fig. 9.9 (b) Again look at the pictures of two mountains given below. Here we notice that the mountain in Fig. 9.0 (a) is more steep compaired to mountain in Fig. 9.0 (b). 0 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

15 (a) Fig. 9.0 (b) How can we quantif this steepness? Here we sa that the angle of inclination of mountain (a) is more than the angle of inclination of mountain (b) with the ground. Tr to see the difference between the ratios of the maimum height from the ground to the base in each case. Naturall, ou will find that the ratio in case (a) is more as compaired to the ratio in case (b). That means we are concerned with height and base and their ratio is linked with tangent of an angle, so mathematicall this ratio or the tangent of the inclination is termed asslope. We define the slope as tangent of an angle. The slope of a line is the the tangent of the angle θ(sa) which the line makes with the positive direction of -ais. Generall, it is denoted b m ( tan θ) Note : If a line makes an angle of 90 0 or 70 0 with the ais, the slope of the line can not be defined. Eample 9. In Fig. 9.9 find the slope of lines AB and BA. Solution : Slope of line AB tan α Slope of line BA tan ( π +α ) tan α. Note : From this eample, we can observe that "slope is independent of the direction of the line segement". Eample 9. Find the slope of a line which makes an angle of 0 0 with the negative direction of -ais. Solution : Here θ m slope of the line tan ( ) tan 0 0 MATHEMATICS ' B ' Fig. 9. A Get Discount Coupons for our Coaching institute and FREE Stud Material at

16 Eample 9.4 Find the slope of a line which makes an angle of 60 0 with the positive direction of -ais. Solution : Here θ m slope of the line tan ( ) ' B A cot 60 0 tan 0 0 Eample : 9.5 If a line is equall inclined to the aes, show that its slope is ±. Solution : Let a line AB be equall inclined to the aes and meeting aes at pointsa and B as shown in the Fig. 9. ' A ' B (a) ' Fig. 9. In Fig 9.(a), inclination of line AB 0 XAB 45 Slope of the line AB tan 45 0 In Fig. 9. (b) inclination of line AB XAB ' B A Slope of the line AB tan5 0 tan ( ) tan 45 0 Thus, if a line is equall inclined to the aes, then the slope of the line will be ±. (b) ' Fig. 9. MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

17 CHECK YOUR PROGRESS 9.5. Find the Slope of a line which makes an angle of (i) 60 0, (ii)50 0 with the positive direction of -ais.. Find the slope of a line which makes an angle of 0 0 with the positive direction of-ais.. Find the slope of a line which makes an angle of 60 0 with the negative direction of-ais. 9.7 SLOPE OF A LINE JOINING TWO DISTINCT POINTS Let A, ) and B, ) be two distinct points. Draw a line through A and B and let the ( ( inclination of this line be θ. Let the point of intersection of a horizontal line througha and a vertical line through B be M, then the coordinates of M are as shown in the Fig. 9.4 (A) ' O A(, ) ' θ (a) B (, ) (, ) M ' Fig. 9.4 O ' M (, ) B (, ) (b) A(, ) In Fig 9.4 (a), angle of inclination MAB is equal to θ (acute). Consequentl. tan θ MB tan( MAB) AM (B) In Fig. 9.4 (b), angle of inclination θ is obtuse, and since θ and MAB are supplementar, consequentl, tan θ tan( MAB) MB MA Hence in both the cases, the slope m of a line through A(,, ) and B, ) is given b ( θ m MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

18 Note : if, then m is not defined. In that case the line is parallel to -ais. Is there a line whose slope is? Yes, when a line is inclined at 45 0 with the positive direction of -ais. Is there a line whose slope is of -ais.? Yes, when a line is inclined at 60 0 with the positive direction From the answers to these questions, ou must have realised that given an real numberm, there will be a line whose slope is m (because we can alwas find an angle α such that tan α m). Eample 9.6 Find the slope of the line joining the points A(6, ) and B(4, 0). Solution : The slope of the line passing through the points (, ) and (, ) Here,, ; 4, Now substituting these values, we have slope 4 6 Eample 9.7 Determine, so that the slope of the line passing through the points (, 6) and (, 4) is. Solution : Slope or CHECK YOUR PROGRESS (Given). What is the slope of the line joining the points A(6, 8) and B(4, 4)?. Determine so that 4 is the slope of the line through the points A(6,) and B(, 8).. Determine, if the slope of the line joining the pointsa( 8, ) and B(, ) is A(, ) B(0, 4) and C( 5, 0) are the vertices of a triangle ABC. Find the slope of the line passing through the point B and the mid point of AC 4 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

19 5. A(, 7), B(, 0), C(4, ) and D(, ) are the vertices of a quadrilateralabcd. Show that (i) slope of AB slope of CD (ii) slope of BC slope of AD 9.8 CONDITIONS FOR PARALLELISM AND PERPENDICULARITY OF LINES Slope of Parallel Lines Let l, l, be two (non-vertical) lines with their slopes m and m respectivel. Let θ and θ be the angles of inclination of these lines respectivel. Case I : Let the lines l and l be parallel Then θ θ tan θ tanθ m m Thus, if two lines are parallel then their slopes are equal. Case II : Let the lines l, and l have equal slopes. i.e. m m tan θ tanθ (0 80 ) l l Hence, two (non-vertical) lines are parallel if and onl if m m 9.8. SLOPES OF PERPENDICULAR LINES Let l and l be two (non-vertical)lines with their slopesm and m respectivel. Also let θ and θ be their inclinations respectivel. l l O θ Fig. 9.5 l θ l θ θ θ θ O O (a) Fig. 9.6 (b) MATHEMATICS 5 Get Discount Coupons for our Coaching institute and FREE Stud Material at

20 Case-I : Let l l 0 0 θ 90 + θ or θ 90 + θ 0 0 tan θ tan( + θ ) or tan θ tan( + θ ) tan θ ( θ ) or tan θ ( θ ) cot cot tan θ tan θ or tan θ tan θ In both the cases, we have tan tan or m.m Thus, if two lines are perpendicular then the product of their slopes is equal to. Case II : Let the two lines l and l be such that the product of their slopes is. i.e. m.m tan tan tan cot tan tan( 90 θ ) 0 ( ) tan Either θ 90 + θ or θ 90 + θ In both cases l l. Hence, two (non-vertical) lines are perpendicular if and onl ifm.m. Eample 9.8 Show that the line passing through the points A(5,6) and B(,) is parallel to the line passing, through the points C(9, ) and D(6, 5). 6 Solution : Slope of the line AB and slope of the line CD 6 9 As the slopes are equal AB CD. 6 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

21 Eample 9.9 Show that the line passing through the points A(, 5) and B(,5) is perpendicular to the line passing through the points L(6,) and M(,). Solution : Here m slope of the line AB and m slope of the line LM Now m.m 5 Hence, the lines are perpendicular to each other. Eample 9.0 Using the concept of slope, show that A(4,4), B(,5) and C are the vertices of a right triangle. 5 4 Solution : Slope of line AB m 4 Slope of line BC m 5 4 and slope of line AC m 4 Now m m AB AC ABC is a right-angled triangle. Hence, A(4,4), B(,5) and C(, ) are the vertices of right triangle. Eample 9. What is the value of so that the line passing through the points A(,) and B(,7) is perpendicular to the line passing through the point C (,4) and D (0,6)? 7 Solution : Slope of the line AB m Slope of the line CD m 0 + Since the lines are perpendicular, MATHEMATICS 7 Get Discount Coupons for our Coaching institute and FREE Stud Material at

22 m m or ( 7) or 4 or or CHECK YOUR PROGRESS 9.7. Show that the line joining the points (, ) and ( 4,) is (i) parallel to the line joining the points (7, ) and (0,). (ii) perpendicular to the line joining the points (4,5) and (0, ).. Find the slope of a line parallel to the line joining the points ( 4,) and (,).. The line joining the points ( 5,7) and (0, ) is perpendicular to the line joining the points (,) and (4,). Find. 4. A(,7), B(,0), C(4,) and D(,) are the vertices of quadrilateral ABCD. Show that the sides of ABCD are parallel. 5. Using the concept of the slope of a line, show that the points A(6, ), B(5,0) and C(,) are collinear.[hint: slopes of AB, BC and CA must be equal.] 6. Find k so that line passing through the points (k,9) and (,7) is parallel to the line passing through the points (, ) and (6,4). 7. Using the concept of slope of a line, show that the points ( 4, ), ( 4), (4,0) and (,) taken in the given order are the vertices of a rectangle. 8. The vertices of a triangle ABC are A(,), B(, 4) and C(5, ). M and N are the midpoints of AB and AC. Show that MN is parallel to BC and MN BC. 9.9 INTERCEPTS MADE BY A LINE ON AXES If a line l (not passing through the Origin) meets -ais at A and -ais at B as shown in Fig. 9.7, then (i) (ii) OA is called the -intercept or the intercept made b the line on -ais. OB is called -intercept or the intercept made b the line on -ais. 8 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

23 (iii) (iv) (v) OA and OB taken together in this order are called the intercepts made b the line l on the aes. AB is called the portion of the line intercepted between the aes. The coordinates of the point A on -ais are (a,0) and those of point B are (0,b) To find the intercept of a line in a given plane on -ais, we put 0 in the given equation of a line and the value of so obtained is called the intercept. To find the intercept of a line on -ais we put 0 and the value of so obtained is called the intercept. Note:. A line which passes through origin makes no intercepts on aes.. A horizontal line has no -intercept and vertical line has no -intercept.. The intercepts on - ais and -ais are usuall denoted b a and b respectivel. But if onl -intercept is considered, then it is usuall denoted b c. Eample 9. If a line is represented b + 6, find its and intercepts. Solution : The given equation of the line is + 6 Putting 0 in (i), we get Thus, -intercept is. Again putting 0 in (i), we get 6 Thus, -intercept is. CHECK YOUR PROGRESS 9.8. Find and intercepts, if the equations of lines are : l O... (i) B b a Fig. 9.7 A (i) + 6 (ii) 7 + (iii) + a b (v) 8 (vi) 7 (iv) a + b c MATHEMATICS 9 Get Discount Coupons for our Coaching institute and FREE Stud Material at

24 9.0 LOCUS OF A POINT 9.0. DEFINITION OF THE LOCUS OF A POINT Locus of a point is the path traced b the point when moving under a given condition or conditions. Thus, locus of a point is a path of definite shape. It ma be a straight line, circle or an other curve. For Eample : (i) The locus of a point in a plane which moves such that it is at constant distance from a fied point in the plane is a circle as shown in Fig. 9.8 (a). (ii) (iii) (a) (b) O c A c O (a) (b) (c) c Fig. 9.8 The locus of a point which moves such that it is alwas at a constant distance from-ais is a pair of straight lines parallel to -ais. [See Fig. 9.8 (b)] The locus of a point in a plane which moves such that it is alwas at a constant distance from the two fied points in the same plane is perpendicular bisector of the line segment joining the two points. [See Fig. 9.8(c)]. From the definition and the eamples of a locus, we observe that Ever point which satisfies the given condition or conditions is a point on the locus. Ever point of the locus must satisf the given condition or conditions EQUATION OF LOCUS The equation of locus of a moving point (,) is an algebraic relation between and satisfing the given conditions of motion of a point. The coordinates (,) of the moving point which generates the locus are called current coordinates. The point covers all the positions on the locus and is called the general point. Let us take an eample : Let P(4,) and Q(7,) be two points. Let us tr to locate a point R which is equidistant from both the points P and Q. Let the s of R be (,). 0 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

25 i. e. P A Get Discount Coupons for our Coaching institute and FREE Stud Material at Then PR ( 4) + ( ) QR ( 7 ) ( + ) PR QR Squaring, we get This is called the equation of the locus of a point R which is equidistant from the points P and Q. From the above, we observe the following working rule: 9.0. WORKING RULE TO FIND THE EQUATION OFTHE LOCUS OFA POINT (i) (ii) (iii) (iv) Take an point (,) on the locus. Write the given geometrical form in the terms of and and known constant or constants and simplif it, if necessar. Epress the given condition in mathematical form in the terms of and and known constant or constants and simplif it, if necessar. The equation so obtained is the equation of the required locus. Eample 9. Find the equation of locus of points which are thrice as far from ( a,0) as from (a,0). Solution : Let P(,) be an point on locus. Also let A( a,0) and B(a,0) be the two given points. Then b the given condition. PA PB 9 PB P(, ) B( a,0) AA(-a, ( a,00) ) B(a, 0) Fig. 9.9 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

26 ( ) ( ) + a + a a+ + a + a + a a + a a 8a 8 0 Thus, + 5a + a 0 is the required equation of the locus. Eample 9.4 Find the equation of the locus of a point such that the sum of its distances from (0,) and (0, ) is 6. P(, ) A(0, ) B(0, -) Fig. 9.0 Solution : Let P(,) be an point on the locus. Also let A(0,) and B(0, ) be the given points. From the given condition, we have PA + PB 6 ( ) + + ( + ) 6 + or Squaring both sides, we get or or Squaring both sides again, we get ( + 9) 9 ( ) MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

27 or or which is the required equation of the locus. Eample 9.5 A(,) and B(,4) are the two vertices of a triangle ABC. Find the equation of the locus of the centroid of the triangle, if the third verte C is a point of the locus whose equation is 4 8. Solution : Let C(a,b) be the third verte of the triangle ABC. Since C(a,b) lies on the locus whose equation is 4 8. a 4b 8 Let (h,k) be the centroid of the triangle ABC. + a b h and k a h and b k 5 Substituting these values of a and b, we get (h ) 4(k 5) 8 h 4k + 0 Hence, the locus of the point G (h,k) is 4+ 0 Eample 9.6 (, ) is a point of the locus whose equation is point of locus, find b. Solution : Since (, ) is a point of the locus whose equation is ( ) a a The equation of the locus is As (8,b) is also a point of this locus.... (i) a. If (8,b) is also a a b 8 b 6 b ± 4 Hence, the value of b ± MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

28 CHECK YOUR PROGRESS 9.9. Find the locus of a point which is equidistant from the points (,4) and ( 4,6).. Find the locus of a point equidistant from the points (4,) and the-ais.. Find the equation of locus of a point which moves so that the distance from the point (4,) is twice its distance from the point (,5). 4. A (,) and B (0,) are the coordinates of the two vertices of a triangle. Find the locus of a point P such that the area of the triangle PAB sq. units. 5. Find the equation of the locus of a point which moves so that the sum of the squares of its distances from the point (,) and (,4) is Find the locus of a point which moves such that the sum of its distances from the points (,0) and (,0) is less than If (h,0) is a point of the locus whose equation is , find h. 8. If, is a point of the locus whose equation is point of the locus. LET US SUM UP a, find b if (b,6) is also a Distance between an two points (, ) and (, ) is ( ) ( ) + s of the point dividing the line segment joining the points (, ) and (, ) internall in the ratio m : m are m + m m + m, m + m m + m s of the point dividing the line segment joining the the points (, ) and ( ) m : m are., eternall are in the ratio m m m m, m m m m s of the mid point of the line segment joining the points (, ) and (, ) are + +, 4 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

29 The area of a triangle with vertices (, ) and (, ) and ( ) [( ) + ( ) + ( )], is given b Three points A, B, and C are collinear if the area of the triangle formed b them is zero. If θ is the angle which a line makes with the positive direction of-ais, then the slope of the line is m tanθ. Slope (m) of the line joining A (, ) and (, ) B is given b m A line with the slope m is parallel to the line with slope m if m m. A line with the slope m is perpendicular to the line with slope m if m m. If a line l (not passing through the origin) meets - ais at A and - ais at B then OA is called the - intercept and OB is called the - intercept. Locus of a point is the path traced b it when moving under given condition or conditions. SUPPORTIVE WEB SITES TERMINAL EXERCISE. Find the distance between the pairs of points: (a) (, 0) and (, cot θ) (b) ( sin A, cosa) and (sin B, cos B). Which of the following sets of points form a triangle? (a) (, ), (, ) and (0, ) (b) (, ), (, ) and (, 0). Find the midpoint of the line segment joining the points (. 5) and ( 6, 8). 4. Find the area of the triangle whose vertices are: (a) (, ), (, ), (, 4) (b)(c, a), (c + a, a), (c a, a) 5. Show that the following sets of points are collinear (b showing that area formed is 0). (a) (, 5) (, ) and (0, ) (b) (a, b + c), (b, c + a) and (c, a + b) MATHEMATICS 5 Get Discount Coupons for our Coaching institute and FREE Stud Material at

30 6. If (, ), (7, 6) and (, a) are collinear, find. 7. Find the area of the quadrilateral whose vertices are (4,) ( 5,6) (0,7) and (, 6). 8. Find the slope of the line through the points (a) (,), (4,) (b) (4, 6), (, 5) 9. What is the value of so that the line pasing through the points (, ) and (,7) is parallel to the line passing through the points (, 4) and (0, 6)? 0. Without using Pthagoras theorem, show that the points (4, 4), (, 5) and (, ) are the vertices of a right-angled triangle.. Using the concept of slope, determine which of the following sets of points are collnear: (i) (, ), (8, 5) and (5, 4), (ii) (5, ), (, ) and (, 4),. If A (, ) and B (, 5) are two vertices of a rectangle ABCD, find the slope of (i) BC (ii) CD (iii) DA.. A quadrilateral has vertices at the points (7, ), (, 0), (0, 4) and (4, ). Using slopes, show that the mid-points of the sides of the quadrilatral form a parallelogram. 4. Find the -intercepts of the following lines: (i) 8 (ii) (iii) 5. Find the equation of the locus of a point equidistant from the points (,4) and-ais. 6. Find the equation of the locus of a point which is equidistant from the points (a+b, a b) and (a b, a+b). 7. Is A(a,0), B( a,0) are two fied points, find the locus of a point P which moves so that PA PB. 8. Find the equation of the locus of a point P if the sum of squares of its distances from (,) and (,4) is 5 units. 6 MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at

31 ANSWERS CHECK YOUR PROGRESS 9. (a) 58 (b) ( a + b ) CHECK YOUR PROGRESS 9.. (a),4 (b) (,5). (,4) 5. (a) (,6) 4.,, 4, CHECK YOUR PROGRESS 9. 5 a. (a) sq. units (b) sq. units (c). 5 k. 80 sq. units 4. CHECK YOUR PROGRESS k 6. k, CHECK YOUR PROGRESS 9.5. (i) (ii) sq. units.. CHECK YOUR PROGRESS CHECK YOUR PROGRESS sq. units k MATHEMATICS 7 Get Discount Coupons for our Coaching institute and FREE Stud Material at

32 CECK YOUR PROGRESS 9.8. (i) -intercept 6 (ii) -intercept 7 (iii) -intercept a -intercept -intercept (iv) -intercept a c -intercept b c -intercept b (v) -intercept 4 (vi) -intercept -intercept 6 -intercept CHECK YOUR PROGRESS < ± 5 8. TERMINAL EXERCISE. (a) cosec θ (b) sin A + B. None of the given sets forms a triangle.., a 7 4. (a) sq. unit (b) a sq. unit sq. unit. 8. (a) 0 (b) 9.. Onl (ii). (i) (ii) 8 (iii) (i) 4 (ii) (iii) a + 5a MATHEMATICS Get Discount Coupons for our Coaching institute and FREE Stud Material at