Module-3-R.C.C.-01-THEORY

1.1 Introduction . Co-ordinates of a point are the real variables associated in an order to a point to describe its location in some space. Here the space is the two dimensional plane. The work of describing the position of a point in a plane by an ordered pair of real numbers can be done in different ways. The two lines XOX' and YOY' divide the plane in four quadrants. XOY, YOX', X' OY', Y'OX are respectively called the first, the second, the third and the fourth quadrants. We assume the directions of OX, OY as positive while the directions of OX', OY' as negative. Quadrant x-coordinate y-coordinate point First quadrant + + (+,+) Second quadrant – + (–,+) Third quadrant – – (–,–) Fourth quadrant + – (+,–) 1.2 Cartesian Co-ordinates of a Point . This is the most popular co-ordinate system. Let us consider two intersecting lines XOX' and YOY', which are perpendicular to each other. Let P be any point in the plane of lines. Draw the rectangle OLPM with its adjacent sides OL,OM along the lines XOX', YOY' respectively. The position of the point P can be fixed in the plane provided the locations as well as the magnitudes of OL, OM are known. Axis of x : The line XOX' is called axis of x. Axis of y : The line YOY' is called axis of y. Co-ordinate axes : x axis and y axis together are called axis of co-ordinates or axis of reference. Origin : The point ‘O’ is called the origin of co-ordinates or the origin. Oblique axes : If both the axes are not perpendicular then they are called as oblique axes. Let OL = x and OM = y which are respectively called the abscissa (or x-coordinate) and the ordinate (or y-coordinate). The co-ordinate of P are (x, y). Note :  Co-ordinates of the origin is (0, 0).  The y co-ordinate of every point on x-axis is zero.  The x co-ordinate of every point on y-axis is zero. 1.3 Polar Co-ordinates . Let OX be any fixed line which is usually called the initial line and O be a fixed point on it. If distance of any point P from the O is 'r' and , then (r, ) are called the polar co-ordinates of a point P. If (x, y) are the cartesian co-ordinates of a point P, then ; and 1.4 Distance Formula . The distance between two points and is given by Note :  The distance of a point from origin .  If distance between two points is given then use sign.  When the line PQ is parallel to the y-axis, the abscissa of point P and Q will be equal i.e, ;  When the segment PQ is parallel to the x-axis, the ordinate of the points P and Q will be equal i.e., . Therefore (1) Distance between two points in polar co-ordinates : Let O be the pole and OX be the initial line. Let P and Q be two given points whose polar co-ordinates are and respectively. Then and then In from cosine rule Note :  Always taking and in radians. Example: 1 If the point (x, y) be equidistant from the points and , then (a) (b) (c) (d) Solution: (d) Let points , A . According to Question, , i.e., = Example: 2 If cartesian co-ordinates of any point are , then its polar co-ordinates is (a) (b) (c) (d) None of these Solution: (c) We know that , , Polar co-ordinates . 1.5 Geometrical Conditions. (1) Properties of triangles (i) In any triangle and . (ii) The is equilateral  . (iii) The is a right angled triangle  or or . (iv) The is isosceles  or or . (2) Properties of quadrilaterals (i) The quadrilateral ABCD is a parallelogram if and only if (a) , or (b) the middle points of BD and AC are the same, In a parallelogram diagonals AC and BD are not equal and . (ii) The quadrilateral is a rectangle if and only if (a) and or, (b) or, (c) the middle points of and BD are the same and AC=BD. ( ) (iii) The quadrilateral is a rhombus (but not a square) if and only if (a) and or, (b) the middle points of AC and BD are the same and but . (iv) The quadrilateral is a square if and only if (a) and or (b) the middle points of AC and BD are the same and , . Note :  Diagonals of square, rhombus, rectangle and parallelogram always bisect each other.  Diagonals of rhombus and square bisect each other at right angle.  Four given points are collinear, if area of quadrilateral is zero. Example: 3 ABC is an isosceles triangle. If the co-ordinates of the base are B(1,3) and C (– 2,7) the co-ordinates of vertex A can be [] (a) (1, 6) (b) (c) (d) None of these Solution: (c) Let the vertex of triangle be . Then the vertex is equidistant from B and C because ABC is an isosceles triangle, therefore =  Thus, any point lying on this line can be the vertex A except the mid point of BC. Hence vertex A is Example: 4 The extremities of diagonal of parallelogram are the points (3, – 4) and (– 6,5) if third vertex is (– 2,1), then fourth vertex is (a) (1,0) (b) (–1,0) (c) (1,1) (d) None of these Solution: (b) Let and be the ends of diagonal of parallelogram ABCD. Let and D be (x, y), then mid points of diagonal AC and BD coincide. So, and .  Coordinates of D are (–1, 0) Example: 5 The vertices A and D of square ABCD lie on positive side of x and y-axis respectively. If the vertex C is the point (12, 17), then the coordinate of vertex B are (a) (14, 16) (b) (15, 3) (c) (17, 5) (d) (17, 12) Solution: (c) Let the co-ordinate of B be (h, k) Draw BL and CM perpendicular to x-axis and y-axis.  and   Hence, Point B is (17, 5). Example: 6 A triangle with vertices (4, 0); (–1, –1); (3, 5) is (a) Isosceles and right angled (b) Isosceles but not right angled (c) Right angled but not isosceles (d) Neither right angled nor isosceles Solution: (a) Let A then , ; i.e. So triangle is isosceles and also . Hence ABC is right angled isosceles triangle. 1.6 Section Formulae. If divides the join of and in the ratio (1) Internal division : If divides the segment AB internally in the ratio of  The co-ordinates of are and (2) External division : If divides the segment AB externally in the ratio of  The co-ordinates of are and Note :  If divides the join of and in the ratio , then . Positive sign is taken for internal division and negative sign is taken for external division.  The mid point of is [Here ]  For finding ratio, use ratio . If is positive, then divides internally and if is negative, then divides externally.  Straight line divides the join of points and in the ratio . If ratio is –ve then divides externally and if ratio is +ve then divides internally. Example: 7 The co-ordinate of the point dividing internally the line joining the points (4, –2) and (8, 6) in the ratio 7: 5 will be (a) (16, 18) (b) (18, 16) (c) (d) Solution: (c) Let point (x, y) divides the line internally. Then = , = . Example: 8 The line divides the line joining the points (–1,1) and (5, 7) in the ratio (a) 2 : 1 (b) 1 : 2 Internally (c) 1 : 2 Externally (d) None of these Solution: (b) Required ratio = = – (Internally) Example: 9 The line joining points (2, –3) and (–5, 6)is divided by y-axis in the ratio (a) 2 : 5 (b) 2 : 3 (c) 3 : 5 (d) 1 : 2 Solution: (a) Let ratio be k : 1 and coordinate of y-axis are (0, b). Therefore, 1.7 Some points of a Triangle. (1) Centroid of a triangle : The centroid of a triangle is the point of intersection of its medians. The centroid divides the medians in the ratio (Vertex : base) If , and are the vertices of a triangle. If G be the centroid upon one of the median (say) AD, then AG : GD = 2 : 1  Co-ordinate of G are Example: 10 The centroid of a triangle is (2,7)and two of its vertices are (4, 8) and (–2, 6) the third vertex is (a) (0, 0) (b) (4, 7) (c) (7, 4) (d) (7, 7) Solution: (b) Let the third vertex (x, y) , i.e. Hence third vertex is (4, 7). (2) Circumcentre : The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of the sides of a triangle. It is the centre of the circle which passes through the vertices of the triangle and so its distance from the vertices of the triangle is the same and this distance is known as the circum-radius of the triangle. Let vertices A, B, C of the triangle ABC be and and let circumcentre be O(x, y) and then (x, y) can be found by solving i.e., Note :  If a triangle is right angle, then its circumcentre is the mid point of hypotenuse.  If angles of triangle i.e., A, B, C and vertices of triangle and are given, then circumcentre of the triangle ABC is Example: 11 If the vertices of a triangle be (2, 1); (5, 2) and (3,4) then its circumcentre is (a) (b) (c) (d) None of these Solution: (b) Let circumcentre be and given points are and  .....(i) and .....(ii) On solving (i) and (ii), we get (3) Incentre : The incentre of a triangle is the point of intersection of internal bisector of the angles. Also it is a centre of a circle touching all the sides of a triangle. Co-ordinates of incentre Where a, b, c are the sides of triangle ABC. (4) Excircle : A circle touches one side outside the triangle and other two extended sides then circle is known as excircle. Let ABC be a triangle then there are three excircles with three excentres. Let opposite to vertices A,B and C respectively. If vertices of triangle are and then , Note :  Angle bisector divides the opposite sides in the ratio of remaining sides e.g.  Incentre divides the angle bisectors in the ratio and  Excentre : Point of intersection of one internal angle bisector and other two external angle bisector is called as excentre. There are three excentres in a triangle. Co-ordinate of each can be obtained by changing the sign of a,b,c respectively in the formula of in-centre. Example: 12 The incentre of the triangle with vertices and is (a) (b) (c) (d) Solution: (d) Here The triangle is equilateral . So, the incentre is the same as the centroid.  Incentre = . (5) Orthocentre : It is the point of intersection of perpendiculars drawn from vertices on opposite sides (called altitudes) of a triangle and can be obtained by solving the equation of any two altitudes. Here O is the orthocentre since , and then Solving any two we can get coordinate of O. Note :  If a triangle is right angled triangle, then orthocentre is the point where right angle is formed.  If the triangle is equilateral then centroid, incentre, orthocentre, circum-centre coincides.  Orthocentre, centroid and circum-centre are always collinear and centroid divides the line joining orthocentre and circum-centre in the ratio 2 : 1  In an isosceles triangle centroid, orthocentre, incentre, circum-centre lie on the same line. Example: 13 The vertices of triangle are (0, 3) (– 3, 0) and (3, 0). The co-ordinate of its orthocentre are (a) (0, – 2) (b) (0, 2) (c) (0, 3) (d) (0, –3) Solution: (c) Here . In a right angled triangle, orthocentre is the point where right angle is formed.  Orthocentre is (0, 3) Example: 14 If the centroid and circumcentre of triangle are (3, 3); (6, 2), then the orthocentre is (a) (9, 5) (b) (3, –1) (c) (– 3, 1) (d) (– 3, 5) Solution: (d) Let orthocentre be . We know that centroid divides the line joining orthocentre and circumcentre in the ratio 2 : 1  , Hence orthocentre is (–3, 5). 1.8 Area of some Geometrical figures. (1) Area of a triangle : The area of a triangle ABC with vertices and . The area of triangle ABC is denoted by ‘’and is given as In equilateral triangle (i) Having sides a, area is . (ii) Having length of perpendicular as 'p' area is . Note :  If a triangle has polar co-ordinates and then its area  If area is a rational number. Then the triangle cannot be equilateral. (2) Collinear points : Three points are collinear. If area of triangle is zero, i.e., (i)  (ii) or or (3) Area of a quadrilateral : If and are vertices of a quadrilateral, then its Area Note :  If two opposite vertex of rectangle are and , then its area is .  It two opposite vertex of a square are and , then its area is (4) Area of polygon : The area of polygon whose vertices are is Or Stair method : Repeat first co-ordinates one time in last for down arrow use positive sign and for up arrow use negative sign.  Area of polygon = Example: 15 The area of the triangle formed by the points is (a) abc (b) (c) (d) 0 Solution: (d) Area = = , (Applying ) = =0 Example: 16 Three points are A(6, 3), B(– 3, 5), C(4, – 2) and P (x, y) is a point, then the ratio of area of and is (a) (b) (c) (d) None of these Solution: (a) = Example: 17 If the points (2K, K)(K, 2K) and (K, K) with enclose triangle of area 18 square units then the centroid of triangle is equal to (a) (8, 8) (b) (4, 4) (c) (– 4, – 4) (d) Solution: (a)   . Consider K = +6 because K > 0, then the points (12, 6) (6,12) and (6,6). Hence, centroid = Example: 18 If the points (x+1, 2); (1,x + 2); are collinear, then x is (a) 4 (b) 0 (c) – 4 (d) None of these Solution: (b,c) Let ; ; . A, B, C are collinear, if area of      Example: 19 The points (1, 1); are collinear for (a) (b) (c) (d) None of these Solution: (b) The given points are collinear, if Area of     Therefore the points are collinear for all value of , except only because at , (Not defined). Example: 20 The points (0, 8/3) (1, 3) and (82, 30) are the vertices of (a) An equilateral triangle (b) An isosceles triangle (c) A right angled triangle (d) None of these Solution: (d) Here , = , Since .  Points A, B, C are collinear. 1.9 Transformation of Axes . (1) Shifting of origin without rotation of axes : Let with respect to axes OX and OY. Let with respect to axes OX and OY and let with respect to axes O'X' and O'Y', where OX and O'X' are parallel and OY and O'Y' are parallel. Then or Thus if origin is shifted to point without rotation of axes, then new equation of curve can be obtained by putting in place of x and in place of y. (2) Rotation of axes without changing the origin : Let O be the origin. Let with respect to axes OX and OY and let with respect to axes OX and OY where then and The above relation between and can be easily obtained with the help of following table (3) Change of origin and rotation of axes : If origin is changed to and axes are rotated about the new origin by an angle in the anticlock-wise sense such that the new co-ordinates of become then the equations of transformation will be and (4) Reflection (Image of a point) : Let be any point, then its image with respect to (i) x axis  (ii) y-axis  (iii) origin  (iv) line  Example: 21 The point (2,3) undergoes the following three transformation successively, (i) Reflection about the line . (ii) Transformation through a distance 2 units along the positive direction of y-axis. (iii) Rotation through an angle of 45o about the origin in the anticlockwise direction. The final coordinates of points are (a) (b) (c) (d) None of these Solution: (b) (i) The new position after reflection is (3,2) (ii) After transformation, it is (3, 2+ 2), i.e, (3, 4) (iii) Rotation makes it , i.e. Example: 22 Reflecting the point (2, –1) about y-axis, coordinate axes are rotated at angle in negative direction without shifting the origin. The new coordinates of the point are (a) (b) (c) (d) None of these Solution: (a) The new position after reflection is (–2, –1) Rotation makes it , i.e., Example: 23 The point (3, 2) is reflected in the y-axis and then moved a distance 5 units towards the negative side of y-axis. The co-ordinate of the point thus obtained are (a) (3, –3) (b) (–3, 3) (c) (3, 3) (d) (¬–3, –3) Solution: (d) Reflection in the y-axis of the point (3,2) is (–3, 2) when it moves towards the negative side of y- axis through 5 units, then the new position is (–3, 2– 5) =(– 3, – 3) 1.10 Locus. Locus : The curve described by a point which moves under given condition or conditions is called its locus. Equation to the locus of a point : The equation to the locus of a point is the relation, which is satisfied by the coordinates of every point on the locus of the point. Algorithm to find the locus of a point Step I : Assume the coordinates of the point say (h, k) whose locus is to be found. Step II : Write the given condition in mathematical form involving h , k. Step III : Eliminate the variable (s), if any. Step IV : Replace h by x and k by y in the result obtained in step III. The equation so obtained is the locus of the point which moves under some stated condition (s) Note :  Locus of a point P which is equidistant from the two point A and B is a straight line and is a perpendicular bisector of line AB.  In above case if PA = kPB where , then the locus of P is a circle.  Locus of P if A and B is fixed. (a) Circle, if = constant (b) Circle with diameter , if (c) Ellipse, if PA +PB = constant (d) Hyperbola, if PA – PB = constant Example: 24 Let A (2, – 3) and B( – 2, 1) be vertices of triangle ABC. If the centroid of this triangle moves on the line , then the locus of the vertex C is the line (a) (b) (c) (d) Solution: (d) Let third vertex C be Centroid = , i.e. According to question,   Hence, locus of vertex C is . Example: 25 The ends of a rod of length l move on two mutually perpendicular lines. The locus of the point on the rod which divides it in the ratio 1 : 2 is (a) (b) (c) (d) None of these Solution: (c) , then or Similarly b = 3k Now we have  Hence locus of P (h, k) is given by Example: 26 If A and B are two fixed points and P is a variable point such that , then the locus of P is a/an (a) Parabola (b) Ellipse (c) Hyperbola (d) None of these Solution: (b) We know that, = constant. Then locus of P is an ellipse.