Chapter-3-PAIR OF STRAIGHT LINE-01- (THEORY)

3.1 Equation of Pair of Straight lines . Let the equation of two lines be .....(i) and .....(ii) Hence is called the joint equation of lines (i) and (ii) and conversely, if joint equation of two lines be then their separate equation will be and . (1) Equation of a pair of straight lines passing through origin : The equation represents a pair of straight line passing through the origin where a, h, b are constants. Let the lines represented by be and where, and = then, and Then, two straight lines represented by are = 0 and . Note :  The lines are real and distinct if  The lines are real and coincident if  The lines are imaginary if  If the pair of straight lines and should have one line common, then .  The equation of the pair of straight lines passing through origin and perpendicular to the pair of straight lines represented by is given by  If the slope of one of the lines represented by the equation be the square of the other, then .  If the slope of one of the lines represented by the equation be times that of the other, then . (2) General equation of a pair of straight lines : An equation of the form, where a, b, c, f, g, h are constants, is said to be a general equation of second degree in x and y. The necessary and sufficient condition for to represent a pair of straight lines is that or (3) Separate equations from joint equation: The general equation of second degree be . To find the lines represented by this equation we proceed as follows : Step I : Factorize the homogeneous part into two linear factors. Let the linear factors be and . Step II: Add constants and in the factors obtained in step I to obtain and . Let the lines be and . Step III : Obtain the joint equation of the lines in step II and compare the coefficients of x, y and constant terms to obtain equations in c' and c" . Step IV : Solve the equations in c' and c" to obtain the values of c' and c". Step V : Substitute the values of c' and c" in lines in step II to obtain the required lines. Example: 1 If the sum of the slopes of the lines given by is four times their product. Then c has the value (a) – 2 (b) – 1 (c) 2 (d) 1 Solution: (c) We know that, and . Given,  Example: 2 If one of the lines represented by the equation be then (a) (b) (c) (d) Solution: (a) Substituting the value of y in the equation  Example: 3 If the equation represent two straight lines, then the value of K is (a) 1 (b) 2 (c) 0 (d) 3 Solution: (b) Condition for pair of lines, , Here Then, . On solving, we get K= 2. 3.2 Angle between the Pair of Lines. (1) The angle between the pair of lines represented by is given by (i) The lines are coincident if the angle between them is zero.  Lines are coincident i.e.,     Hence, the lines represented by are coincident, (ii) The lines are perpendicular if the angle between them is .       coeff. of coeff. of Thus, the lines represented by are perpendicular iff i.e., coeff. of coeff. of . (2) The angle between the lines represented by is given by (i) The lines are parallel if the angle between them is zero. Thus, the lines are parallel iff    . Hence, the lines represented by are parallel iff and or . (ii) The lines are perpendicular if the angle between them is . Thus, the lines are perpendicular i.e.,     coeff. of coeff. of Hence, the lines represented by are perpendicular iff i.e., coeff. of coeff. of . (iii) The lines are coincident, if . Example: 4 The angle between the lines is (a) (b) (c) (d) Solution: (b) Angle between the lines is = , Example: 5 If the angle between the pair of straight lines represented by the equation is , where is a non- negative real number, then is (a) 2 (b) 0 (c) 3 (d) 1 Solution: (a) Given that Now, since  =       or – 40, but is a non-negative real number. Hence . Example: 6 The angle between the pair of straight lines represented by is (a) (b) (c) (d) Solution: (b) Angle between the lines is ,   3.3 Bisectors of the Angles between the Lines. (1) The joint equation of the bisectors of the angles between the lines represented by the equation is .....(i) Here, coefficient of coefficient of . Hence, the bisectors of the angles between the lines are perpendicular to each other. The bisector lines will pass through origin also. Note :  If , the bisectors are i.e.,  If , the bisectors are i.e., .  If bisectors of the angles between lines represented by and are same, then .  If the equation has one line as the bisector of the angle between the coordinate axes, then . (2) The equation of the bisectors of the angles between the lines represented by + are given by , where ,  is the point of intersection of the lines represented by the given equation. Example: 7 The equation of the bisectors of the angles between the lines represented by is (a) (b) (c) (d) None of these Solution: (a) Equation of bisectors is given by or  Example: 8 If the bisectors of the lines be then (a) (b) (c) (d) Solution: (a) Bisectors of the angle between the lines is  But it is represented by . Therefore 3.4 Point of Intersection of Lines represented by ax2+2hxy+by2+2gx+2fy+c = 0. Let (Keeping y as constant) and (Keeping x as constant) For point of intersection and We obtain, and On solving these equations, we get i.e. Also, since , from first two rows a h g  and h b f  and then solve, we get the point of intersection. Note :  The point of intersection of lines represented by is (0, 0). Example: 9 The point of intersection of the lines represented by the equation is (a) (b) (c) (d) Solution: (c) Let and On solving these equations, we get Trick : If the equation is The points of intersection are given by . Hence point is (– 2, 0) Example: 10 If the pair of straight lines and line are concurrent, then a = (a) – 1 (b) 0 (c) 3 (d) 1 Solution: (d) Given that equation of pair of straight lines   or The intersection point of is (1,1)  Lines and are concurrent.  The intersecting points of first two lines satisfy the third line. Hence,  3.5 Equation of the Lines joining the Origin to the Points of Intersection of a given Line and a given Curve . The equation of the lines which joins origin to the point of intersection of the line and curve , can be obtained by making the curve homogeneous with the help of line , which is We have ......(i) and .....(ii) Suppose the line (ii) intersects the curve (i) at two points A and B. We wish to find the combined equation of the straight lines OA and OB. Clearly OA and OB pass through the origin, so their joint equation is a homogeneous equation of second degree in x and y. From equation (ii), ......(iii) Now, consider the equation .....(iv) Clearly, this equation is a homogeneous equation of second degree. So, it represents a pair of straight lines passing through the origin. Moreover, it is satisfied by the points A and B. Hence (iv) represents a pair of straight lines OA and OB through the origin O and the points A and B which are points of intersection of (i) and (ii). Example: 11 The lines joining the origin to the point of intersection of the circle and the line are (a) (b) (c) (d) Solution: (a,b,c,d) Make homogenous the equation of circle, we get Hence, the equation are and Also after rationalizing these equations becomes and . Example: 12 The pair of straight lines joining the origin to the points of intersection of the line and the circle are at right angles, if (a) (b) (c) (d) Solution: (c) Pair of straight lines joining the origin to the points of intersection of the line and the circle are    If these lines are perpendicular,  . 3.6 Removal of First degree Terms. Let point of intersection of lines represented by ......(i) is . Here For removal of first degree terms, shift the origin to i.e., replacing x by and y be in (i). Alternative Method : Direct equation after removal of first degree terms is Where and 3.7 Removal of the Term xy from f (x, y) = ax2 + 2hxy +by2 without changing the Origin . Clearly, . Rotating the axes through an angle , we have, and After rotation, new equation is Now coefficient of XY = 0. Then we get cot Note :  Usually, we use the formula, for finding the angle of rotation, . However, if , we use as in this case is not defined. Example: 13 The new equation of curve after removing the first degree terms (a) (b) (c) (d) None of these Solution: (c) Let .....(i)  and Their point of intersection is Here Shift the origin to (1, –1) then replacing and in (i), the required equation is i.e., Alternative Method : Here and and Removed equation is i.e., . Example: 14 Mixed term xy is to be removed from the general equation , one should rotate the axes through an angle given by = (a) (b) (c) (d) Solution: (d) Let be the coordinates on new axes, then put in the equation, then the coefficient of xy in the transformed equation is 0. So,  3.8 Distance between the Pair of parallel Straight lines s If represent a pair of parallel straight lines, then the distance between them is given by or Example: 15 Distance between the pair of lines represented by the equation (a) (b) (c) (d) Solution: (c) The distance between the pair of straight lines given by is , Here Example: 16 Distance between the lines represented by the equation is (a) 5/2 (b) 5/4 (c) 5 (d) 0 Solution: (a) First check for parallel lines i.e., which is true, hence lines are parallel. Distance between them is 3.9 Some Important Results (1) The lines joining the origin to the points of intersection of the curves and will be mutually perpendicular, if . (2) If the equation represents a pair of straight lines, then . (3) The pair of lines with the line form an equilateral triangle. (4) The area of a triangle formed by the lines and is given by (5) The lines joining the origin to the points of intersection of line and the circle will be mutually perpendicular, if . (6) If the distance of two lines passing through origin from the point is d, then the equation of lines is (7) The lines represented by the equation will be equidistant from the origin, if (8) The product of the perpendiculars drawn from on the lines is given by (9) The product of the perpendiculars drawn from origin on the lines is (10) If the lines represented by the general equation are perpendicular, then the square of distance between the point of intersection and origin is (11) The square of distance between the point of intersection of the lines represented by the equation and origin is Example: 17 The area of the triangle formed by the lines and is (a) 2 (b) 3 (c) (d) Solution: (c) The area of triangle formed by the lines and is given by Here , then area of triangle = = Example: 18 The orthocentre of the triangle formed by the lines and is (a) (0, 0) (b) (c) (d) Solution: (a) Lines represented by is , . Then the triangle formed is right angled triangle at O(0, 0), therefore O(0, 0) is its orthocentre. Example: 19 If the pair of straight lines given by forms an equilateral triangle with line then is (a) (b) (c) (d) Solution: (d) We know that the pair of lines with the line form an equilateral triangle. Hence comparing with then , Now  .