POINT Preface The concept of Point is ver important for the stud of coordinate geometr. This chapter deals with various forms of representing a Point and several associated properties. The concept of coordinates and basics of trigonometr are required to stud this chapter. This book consists of theoretical and practical eplanations of all the concepts involved in the chapter. Each article is followed b a ladder of illustration. At the end of the theor part there are miscellaneous solved eamples which involve the application of multiple concepts of this chapter. Students are advised to go through all these solved eamples in order to develop better understanding of the chapter and to have better grasping level in the class. Total No. of questions in Point are : In Chapter Eamples...8 Solved Eamples... Total no. of questions... 40 IIT JEE PREPARATION POINT
. SYSTEM OF CO-ORDINATES. Cartesian Co-ordinates : Let XOX and YOY be two perpendicular straight lines drawn through an point O in the plane of the paper. Then.. Ais of : The line XOX is called ais of... Ais of : The line YOY is called ais of... Co-ordinate aes : ais and ais together are called ais of co-ordinates or ais of reference...4 Origin : The point O is called the origin of co-ordinates or the Origin...5 Oblique ais : If both the aes are not perpendicular then the are called as Oblique aes...6 Cartesian Co-ordinates : The ordered pair of perpendicular distance from both ais of a point P ling in the plane is called Cartesian Co-ordinates of P. If the Cartesian co-ordinates of a point P are ( ) then is called abscissa or coordinate of P and is called the ordinate or co-ordinate of point P. Y P() X O X Y (i) Co-ordinates of the origin is (0 0). (ii) co-ordinate on - ais is zero. (iii) co-ordinate on - ais is zero.. Polar Co-ordinates : Let OX be an fied line which is usuall called the initial line and O be a fied point on it. If distance of an point P from the pole O is r and XOP then (r ) are called the polar co-ordinates of a point P. If ( ) are the Cartesian co-ordinates of a point P then P(r) r O and r cos ; r r sin tan Eample based on Sstem of Co-ordinates E. If Cartesian co-ordinates of an point are ( ) then its polar co-ordinates is - (A) ( /) (B) ( /6) (C) ( /6) r r cos r sin ( ). tan / 6 ( /6) E. If polar coordinates of an points are ( /) then its Cartesian coordinates is - (A) ( ) (B) ( ) (C) ( ) cos / sin / ; ( ) Ans. [B]. DISTANCE FORMULA The distance between two points P( ) and Q ( ) is given b PQ ) ( ) ( (i) Distance of a point P() from the origin (ii) Distance between two polar co-ordinates A(r ) and B(r ) is given b AB r r rr cos( ) Eample based on Distance Formula E. Find the distance between P( ) and Q( 7 5). PQ ( 7) ( 5) IIT JEE PREPARATION POINT
E.4 E.5 00 9 09 Ans. Distance of a point P(8 6) from origin is 8 6 00 0 Find the distance between P( ) and Q ( ) 6 6 PQ...cos( ) 6 6 E.6 D E.7 cos( ) 7 Ans. Distance between points (a 0) and (0 a) is (A) a (B) a (C) a (D) a ( a 0) (0 a) a a a a Ans. [A] If distance between the point ( ) and ( 4) is then the value of is - (A) (B) (C) (D) 0 ( ) ( 4) ( ) 4 Squaring both sides 4 ( ) + 4 0 E.8 The point whose abscissa is equal to its ordinate and which is equidistant from the point (0) and B(0) is - (A) ( ) (B) ( ) (C) () (D) (4 4) Let the point P(k k) given PA PB ( k ) k k (k ) k k + k 6k + 9 4k 8 k Ans. [B]. APPLICATIONS OF DISTANCE FORMULA. Position of Three Points : Three given points A B C are collinear when sum of an two distance out of AB BC CA is equal to remaining third distance. Otherwise the points will be vertices of a triangles. (b) Isosceles triangle when an two distance are equal. (c) Right angle triangle when sum of square of an two distances is equal to square of the third distance.. Position of four Points : Four given point A B C and D are vertices of a (a) Square if AB BC CD DA and AC BD (b) Rhombus if AB BC CD DA & AC BD (c) Parallelogram if AB DC; BC AD; AC BD (d) Rectangle if AB CD; BC DA; AC BD Quadrilateral Diagonals Angle between diagonals (i) Parallelogram Not equal (ii) Rectangle Equal (iii) Rhombus Not equal (iv) Square Equal (i) Diagonal of square rhombus rectangle and parallelogram alwas bisect each other. (ii) Diagonal of rhombus and square bisect each other at right angle. (iii) Four given points are collinear if area of quadrilateral is zero. Eample based on Applications of Distance Formula CA 8 68.. Tpes of Triangle : If A B and C are vertices of as AB + BC 4 + 4 CA and triangle then it would be. (a) Equilateral triangle when AB BC CA. IIT JEE PREPARATION POINT E.9 AB The point A(8); B(5 ) and C(0 0) are vertices of (A) An equilateral triangle (B) A right angled triangle (C) An isosceles right angled triangle (D) An isosceles triangle BC ( 8 5) ( ) 4 5 ( ) 4
E.0 AB AB BC Hence the given vertices are of an isosceles right angled triangle. Show that the points A() B( 7) and C( ) are collinear BC CA ( ) ( 7) 9 6 5 ( 5) (7 ) 5 00 5 5 ( ) ( ) 4 6 5 Clearl BC AB + AC. Hence A B C are collinear 4. SECTION FORMULA Co-ordinates of a point which divides the line segment joining two points P( ) and Q( ) in the ratio m : m are. (i) For internal division m m m m (ii) For eternal division m m m m m m m m m m m m (iii) Co-ordinates of mid point of PQ are put m m ; (i) Co-ordinates of an point on the line segment joining two points P( ) and Q( ) are ( ) (ii) Lines joins ( ) and ( ) is divided b (a) ais in the ratio / (b) ais in the ratio / if ratio is positive divides internall if ratio is negative divides eternall. (iii) Line a + b + c 0 divides the line joining the points ( ) and ( ) in the ratio Eample based on Section Formula E. a a b b c c Find the co ordinates of point of internal and eternal division of the line segment joining two points ( ) and ( 4) in the ratio : Internal division E. E. () () (4) ( ) Hence point ( ) Ans. () () Eternal division Hence point ( ) Ans. Mid points of ( ) and (6 7) is 6 7 (4 5) Ans. Ratio in which the line + 4 7 divides the line segment joining the points () and ( ) is - () 4() 7 ( ) 4() 7 E.4 E.5 4 9 4 9 Ans. The points of trisection of line joining the points A ( ) and B (5 ) are 5 7 7 5 (A) 4 (B) 4 5 7 7 7 (C) 4 (D) 4.. () (5) A P P. 5 5 P () 5 7 P () 4. The ratio in which the lines joining the ( 4) and ( 5 6) is divided b -ais (A) : (B) 6 : 4 (C) : (D) None 4 : Ans. [A] 6 5. CO-ORDINATE OF SOME PARTICULAR IIT JEE PREPARATION POINT 4
POINTS Let A( ) B( ) and C( ) are vertices of an triangle ABC then 5. Centroid : The centroid is the point of intersection of the medians (Line joining the mid point of sides and opposite vertices). A can be obtained b changing the sign of a b c respectivel in the formula of In centre. 5. Circumcentre : It is the point of intersection of perpendicular bisectors of the sides of a triangle. It is also the centre of a circle passing vertices of the triangle. If O is the circumcentre of an triangle ABC then OA OB OC ( A ) B F D G Centroid divides the median in the ratio of :. Coordinates of centroid G E C 5. Incentre : The incentre of the point of intersection of internal bisector of the angle. Also it is a centre of a circle touching all the sides of a triangle. A ( ) B D O E ( C ) If a triangle is right angle then its circumcentre is the mid point of hpotenuse. 5.4 Ortho Centre : It is the point of intersection of perpendicular drawn from vertices on opposite sides (called altitudes) of a triangle and can be obtained b solving the equation of an two altitudes. ( A ) B F D Co-ordinates of incentre a b c a b c where a b c are a b c a b c the sides of triangle ABC. (i) Angle bisector divides the opposite sides in the ratio of remaining sides eg. BD DC AB AC c b (ii) Incentre divides the angle bisectors in the ratio (b + c):a (c + a):b and (a + b):c (iii) Ecentre : Point of intersection of one internal angle bisector and other two eternal angle bisector is called as ecentre. There are three ecentre in a triangle. Co-ordinate of each E C ( ) B IIT JEE PREPARATION POINT 5 D O E ( C ) If a triangle is right angle triangle then orthocentre is the point where right angle is formed. Remarks : (i) If the triangle is equilateral then centroid incentre orthocentre circumcentre coincides (ii) Ortho centre centroid and circumcentre are alwas colinear and centroid divides the line joining orthocentre and circumcentre in the ratio : (iii) In an isosceles triangle centroid orthocentre incentre circumcentre lies on the same line. Eample based on Coordinates of Some Particular Points E.6 Centroid of the triangle whose vertices are (0 0) ( 5) and (7 4) is 0 7 0 5 4 ( ) Ans.
E.7 Incentre of triangle whose vertices are A ( 6 7) B (0 7) C (0 8) is - Using distance formula a BC b CA c AB 0 (7 8) 5 6 (7 8) 9 ( 6 0) (7 7) 56 5( 6) 9(0) 56(0) 5(7) 9(7) 56( 8) I 5 9 56 5 9 56 I ( 0) Ans. E.8 If (4) is the centroid of a triangle and its two vertices are (4 ) and ( 97) then third vertices is - (A) (7 8) (B) (8 7) (C) (8 8) (D) (6 8) Let the third vertices of triangle be ( ) then E.9 4 9 7 4 8 8 If (0 ) ( ) and ( 0) are middle points of the sides of a triangle then its incentre is - (A) ( + ) (B) ( ) (C) ( + + ) (D) ( + ) Let A( ) B( ) and C( ) are vertices of a triangle then + 0 + + + + + 0 Solving these equations we get A(0 0) B(0 ) and C( 0) Now a BC b CA c AB Thus incentre of ABC is ( ) Ans. [B] 6. AREA OF TRIANGLE AND QUADRILATERAL 6. Area of Triangle Let A( ) B( ) and C( ) are vertices of a triangle then - Area of Triangle ABC [ ( ) + ( ) + ( )] (i) If area of a triangle is zero then the points are collinear. (ii) In an equilateral triangle IIT JEE PREPARATION POINT 6 (a) having sides a area is a 4 (b) having length of perpendicular as p area is p (iii) If a triangle has polar co-ordinate (r ) (r ) and (r ) then its area [r r sin( ) + r r sin( ) + r r sin( )] 6. Area of quadrilateral : If ( ) ( ) ( ) and ( 4 4 ) are vertices of a quadrilateral then its area [( ) + ( ) + ( 4 4 ) + ( 4 4 )] (i) If the area of quadrilateral joining four points is zero then those four points are colinear. (ii) If two opposite verte of rectangle are ( ) and ( ) and sides are parallel to coordinate aes then its area is ( ) ( ) (iii) If two opposite verte of a square are A ( ) and C ( ) then its area is AC [( ) + ( ) ] Eample based on Area of Triangle & Quadrilateral E.0 If the vertices of a triangle are ( ) (4 6) and ( 5) then its area is - [( 6 5) + 4(5 ) + ( + 6)] [ + + 4]
5 square unit Ans. Y Y P() X O X E. E. If () (4) (5 ) and (4 7) are vertices of a quadrilateral then its area [ 4 + ( ) 5(4) + 5( 7) 4( ) + 4() ( 7)] [4 6 0 5 + 8 + 4 + 7] 4 units. Ans. If the coordinates of two opposite verte of a square are (a b) and (b a) then area of square is - (A) (a b) (B) a + b (C) (a b) (D) (a + b) We know that Area of square d [(a b) + (b a) ] (a b) Ans. [A] 7. TRANSFORMATION OF AXES 7. Parallel transformation : Let origin O(0 0) be shifted to a point (a b) b moving the ais and ais parallel to themselves. If the co-ordinate of point P with reference to old ais are ( ) then co-ordinate of this point with respect to new ais will be ( a b) P( ) P( a b) Y (0 0) O O (ab) P( ) 7. Rotational transformation : Let OX and OY be the old ais and OX and OY be the new ais obtained b rotating the old OX and OY through an angle. X Again if co-ordinates of an point P( ) with reference to new ais will be ( ) then cos + sin sin + cos cos sin sin + cos The above relation between ( ) and ( ) can be easil obtained with the help of following table. ' ' cos sin sin cos 7. Reflection (Image) of a Point : Let ( ) be an point then its image w.r.t. (i) -ais ( ) (ii) -ais ( ) (iii) origin ( ) (iv) line ( ) Eample based on Transformation of Aes E. If ais are transformed from origin to the point ( ) then new co-ordinates of (4 5) is - (A) (6 4) (B) ( 6) (C) (6 6) (D) ( 4) [4 ( ) 5 ] (6 6) E.4 Keeping the origin constant ais are rotated at an angle 0º in negative direction then coordinate of () with respect to old ais is - (A) (B) (C) IIT JEE PREPARATION POINT 7
E.5 cos( 0º ) sin( 0º ) sin( 0º ) cos( 0º ) cos0º + sin0º sin0º + cos0º Ans. [B] If the new coordinates of a point after the rotation of ais in the negative direction b an angle of / are (4 ) then coordinate with respect to old ais are - (A) ( + + ) (B) ( + ) (C) ( + + ) (D) ( ) 4 cos( 60º ) sin( 60º ) sin( 60º ) cos( 60º ) cos sin 8. LOCUS 4 cos ( 60) sin( 60) + sin cos 4 sin ( 60) + cos( 60) + Hence the coordinates are ( + + ). A locus is the curve traced out b a point which moves under certain geometrical conditions. To find a locus of a point first we assume the Co-ordinates of the moving point as (h k) then tr to find a relation between h and k with the help of the given conditions of the problem. In the last we replace h b and k b and get the locus of the point which will be an equated between and. (i) Locus of a point P which is equidistant from the two point A and B is straight line and is a perpendicular bisector of line AB. (ii) In above case if PA kpb where k then the locus of P is a circle. (iii) Locus of P if A and B is fied. (a) Circle if APB Constant (b) Circle with diameter AB if ABB (c) Ellipse if PA + PB Constant (d) Hperbola if PA PB Constant Eample based on Locus E.6 E.7 The locus of a point which is equidistant from point (6 ) and ( ) Let the point is (h k) then ( 6) (k ) h k Hence locus is ( h ) (k ) Ans. Find the locus of a point such that the sum of its distance from the points (0 ) and (0 ) is 6. Let P (h k) be an point on the locus and let A(0) and B(0 ) be the given points. B the given condition PA + PB 6 ( h 0) (k ) + (h 0) (k ) 6 h (k ) 6 h + (k + ) 6 8k 6 h (k ) h + h + (k + ) (k ) (k + 9) 9(h + (k + ) ) E.8 h (k ) 4k + 6k + 8 9h + 9k + 6k + 6 9h + 5k 45 Hence locus of (h k) is 9 + 5 45 Ans. A (a0 ) and B ( a 0) are two fied points of ABC. If its verte C moves in such wa that cota + cotb where is a constant then the locus of the point C is (A) a (B) a (C) a We ma suppose that coordinates of two fied points A B are (a 0) and ( a 0) and variable point C is (h k). From the adjoining figure IIT JEE PREPARATION POINT 8
C(h k) DA a h cot A CD k BD a h cot B CD k B O D A ( a o) (a o) But cot A + cot B so we have a h a h a + k k k Hence locus of C is a 9. SOME IMPORTANT POINTS (i) Ans. [A] Quadrilateral containing two sides parallel is called as Trapezium whose area is given b (sum of parallel sides) (Distance between parallel sides) (ii) A triangle having vertices (at at ) (at at ) and (at at ) then area is a [(t t ) (t t ) (t t )] (iii) Area of triangle formed b Co-ordinate ais and c the line a + b + c 0 is equal to ab (iv) When co-ordinate or co-ordinate of all verte of triangle are equal then its area is zero. (v) In a Triangle ABC of D E F are midpoint of sides AB BC and CA then EF BC and DEF 4 (ABC) B F (vi) Area of Rhombus formed b D A c a ± b ± c 0 is ab (vii) Three points ( ) ( ) ( ) are collinear if (viii)when one verte is origin then area of triangle ( ) (i) To remove the term of in the equation E a + h + b 0 the angle through which the ais must be turned (rotated) is given b tan h a b C IIT JEE PREPARATION POINT 9
E. The point A divides the join of the points ( 5) and (5) in the ratio k : and coordinates of points B and C are ( 5) and (7 ) respectivel. If the area of ABC be units then k equals - (A) 7 9 (B) 6 7 (C) 7 /9 (D) 9 /9 k 5 5k A k k Area of ABC units k 5 5k (5 ) k k 4k 66 ± 4 (k+) SOLVED EXAMPLES 5k + 7 5 ± k k 7 or /9 E. The vertices of a triangle are A(0 6) B ( 6 0) and C () respectivel then coordinates of the e-centre opposite to verte A is - (A) ( / /) (B) ( 4 /) (C) ( / /) (D) ( 4 6) a BC ( 6 ) (0 ) 50 5 b CA ( 0) ( 6) 50 5 c AB (0 6) ( 6 0) 7 6 coordinates of E-centre opposite to verte A are a b c a b c 5.0 5 5 5 4 4 6 a b c a b c 6 ( 6) 6 () E. E.4 5 ( 6) 5.0 6 () 6 6 5 5 6 6 Hence coordinates of e-centre are ( 4 6) Ans. [D] If the middle point of the sides of a triangle ABC are (0 0); ( ) and ( 4) then the area of triangle is (A) 40 (B) 0 (C) 0 (D) 60 If the given mid points be D E F; then the area of DEF is given b [0( 4) + (4 0) (0 )] [0 + 4 + 6] 5 Area of the triangle ABC 4 5 0 Ans. [B] The three vertices of a parallelogram taken in order are ( 0) ( ) and ( ) respectivel. Find the coordinate of the fourth verte - (A) ( ) (B) ( ) (C) ( ) (D) ( ) Let A( 0) B( ) C( ) and D( ) be the vertices of a parallelogram ABCD taken in order. Since the diagonals of a parallelogram bisect each other. Coordinates of the mid point of AC Coordinates of the mid-point of BD 0 and and. Hence the fourth verte of the parallelogram is ( ) Ans. [B] IIT JEE PREPARATION POINT 0
E.5 Which of the following statement is true? (A) The Point A(0 ) B() C(0) and D( ) are vertices of a rhombus (B) The points A( 4 ) B( 4) C(4 0) and D( ) are vertices of a square (C) The points A( ) B( 0) C(4 ) and D( ) are vertices of a parallelogram Here (i) A(0 ) B() C(0) D( ) for a rhombus all four sides are equal but the diagonal are not equal we see AC BD 4 0 4 Hence it is a square not rhombus (ii) Here AB BC 6 4 5 AB BC Hence not square. (iii) In this case mid point of AC is 4 or () 0 4 4 0 Also midpoint of diagonal BD or ( ) Hence the point are vertices of a parallelogram. The students should note that the squares rhombus and the rectangle are also parallelograms but ever parallelogram is not square etc. The desired answer should be pinpointed carefull. E.6 The condition that the three points (a 0) (at at ) and (at at ) are collinear if - (A) t + t 0 (B) t t (C) t t Here the points are collinear if the area of the triangle is zero. Hence / [a(t )at at (at a)] 0 or t (t ) t (t ) 0 t t t t t + t 0 (t t )( t t + ) 0 t t t t + 0 t t The students should note that the points lie on the parabola 4a and (a0) is focus the condition t t is well known condition for E.7 the etremities of a focal chord as we shall see in parabola in our further discussions. If the origin is shifted to ( ) and ais are rotated through an angle of 0º the co-ordinate of () in the new position are (A) (B) (C) If coordinates are ( ) then h + cos sin. k + sin + cos Where ( ) ( ) (h k) ( ) 0º + cos 0 sin 0 0 and + sin0 + cos 0 ' ' Ans. [B] E.8 The locus of the point so that the join of ( 5 ) and ( ) subtends a right angle at the moving point is (A) + + 0 (B) + + 0 (C) + + 0 (D) + 0 Let P (h k) be moving point and let A( 5 ) and B() be given points. E.9 B the given condition APB 90º APB is a right angled triangle. AB AP + PB (+5) + ( ) (h+5) + (k ) + (h ) + (k ) 65 (h + k + h k) + 9 h + k + h k 0 Hence locus of (h k) is + + 0 Ans. [A] The ends of the rod of length moves on two mutuall perpendicular lines find the locus of the point on the rod which divides it in the ratio m : m IIT JEE PREPARATION POINT
(A) m + m ( m m ) (B) (m ) + (m ) mm m m (C) (m ) + (m ) mm m m Let ( ) be the point that divide the rod AB in the ratio m : m and OA a OB b sa a + b...() E.0 m a Now m m b m m Y (0 b) B b m m a m m m m b m m ( ) m O a A(a 0) These putting in () m) m ( m ( m m + m Locus of ( ) is X ) m + m mm m m A point P moves such that the sum of its distance from (ae 0) and ( ae 0)) is alwas a then locus of P is (when 0 < e < ) (A) a a ( e ) Then PA + PB a ( h ae) (k 0) + ( h ae) k a ( h ae) (k 0) a ( h ae) (k 0) Squaring both sides we get (h ae) + k 4a + (h + ae) + k (eh + a) 4aeh 4a 4a 4a ( h ae) k ( h ae) k ( h ae) k [Squaring both sides] (eh + a) (h + ae) + k e h + eah + a h + eah + a e + k h ( e ) + k a ( e ) h k a a ( e ) Hence the locus of (h k) is E. Ans. [A] a a ( e ) The orthocentre of triangle with vertices and is - (A) (C) Here AB BC CA 6 (B) (D) 5 4 ( 4 ) (B) Here BC + CA AB a a ( e ) ABC is right - angled triangle (C) Thus point C is the ortho-centre a ( e ) a Ans. [B] Let P(h k) be the moving point such that the E. The number of points on -ais which are at a sum of its distance from A(ae 0) and B( ae 0) distance c(c < ) from the point ( ) is - (A) (B) is a. IIT JEE PREPARATION POINT
(C) infinite (D) no point Let a point on -ais is ( 0) then its distance from the point ( ) ( ) 9 c or ( ) c 9 9 c But C < c 9 < 0 will be imaginar Ans. [D] IIT JEE PREPARATION POINT