Chapter-2-STRAIGHT LINE-01- (THEORY)

2.1 Definition. The straight line is a curve such that every point on the line segment joining any two points on it lies on it. The simplest locus of a point in a plane is a straight line. A line is determined uniquely by any one of the following: (1) Two different points (because we know the axiom that one and only one straight line passes through two given points) (2) A point and a given direction. Thus, to determine a line uniquely, two geometrical conditions are required. 2.2 Slope (Gradient) of a Line. The trigonometrical tangent of the angle that a line makes with the positive direction of the x-axis in anticlockwise sense is called the slope or gradient of the line. The slope of a line is generally denoted by m. Thus, m = tan  (1) Slope of line parallel to x – axis is . (2) Slope of line parallel to y – axis is . (3) Slope of the line equally inclined with the axes is 1 or –1. (4) Slope of the line through the points and is taken in the same order. (5) Slope of the line is . (6) Slope of two parallel lines are equal. (7) If and be the slopes of two perpendicular lines, then . Note :  m can be defined as for and  If three points A, B, C are collinear, then Slope of AB = Slope of BC = Slope of AC Example: 1 The gradient of the line joining the points on the curve , whose abscissae are 1 and 3, is (a) 6 (b) 5 (c) 4 (d) 3 Solution: (a) The points are (1, 3) and (3, 15) Hence gradient is = Example: 2 Slope of a line which cuts intercepts of equal lengths on the axes is (a) – 1 (b) 0 (c) 2 (d) Solution: (a) Equation of line is   . Hence slope of the line is – 1. 2.3 Equations of Straight line in Different forms. (1) Slope form : Equation of a line through the origin and having slope m is y = mx. (2) One point form or Point slope form : Equation of a line through the point and having slope m is . (3) Slope intercept form : Equation of a line (non-vertical) with slope m and cutting off an intercept c on the y-axis is y = mx + c. The equation of a line with slope m and the x-intercept d is (4) Intercept form : If a straight line cuts x-axis at A and the y-axis at B then OA and OB are known as the intercepts of the line on x-axis and y-axis respectively. The intercepts are positive or negative according as the line meets with positive or negative directions of the coordinate axes. In the figure, OA = x-intercept, OB = y-intercept. Equation of a straight line cutting off intercepts a and b on x–axis and y–axis respectively is . Note :  If given line is parallel to X axis, then X-intercept is undefined.  If given line is parallel to Y axis, then Y-intercept is undefined. (5)Two point form: Equation of the line through the points A and is . In the determinant form it is gives as: = 0 is the equation of line. (6) Normal or perpendicular form : The equation of the straight line upon which the length of the perpendicular from the origin is p and this perpendicular makes an angle  with x-axis is . (7) Symmetrical or parametric or distance form of the line : Equation of a line passing through and making an angle  with the positive direction of x-axis is , where r is the distance between the point P (x, y) and . The coordinates of any point on this line may be taken as , known as parametric co-ordinates, ‘r’ is called the parameter. Note :  Equation of x-axis  y = 0 Equation a line parallel to x-axis (or perpendicular to y-axis) at a distance ‘b’ from it   Equation of y-axis  x = 0 Equation of a line parallel to y-axis (or perpendicular to x-axis) at a distance ‘a’ from it  Example: 3 Equation to the straight line cutting off an intercept 2 from the negative direction of the axis of y and inclined at 30° to the positive direction of x, is (a) (b) (c) (d) Solution: (d) Let the equation of the straight line is . Here and c = – 2 Hence, the required equation is  . Example: 4 The equation of a straight line passing through (– 3, 2) and cutting an intercept equal in magnitude but opposite in sign from the axes is given by (a) (b) (c) (d) Solution: (a) Let the equation be  But it passes through (–3, 2), hence . Hence the equation of straight line is . Example: 5 The equation of the straight line passing through the point (4, 3) and making intercept on the co-ordinates axes whose sum is – 1, is (a) and (b) and (c) and (d) and Solution: (d) Let the equation of line is , which passes through (4, 3). Then  Hence equation is and . Example: 6 Let PS be the median of the triangle with vertices and R(7, 3). The equation of the line passing through (1, – 1) and parallel to PS is (a) (b) (c) (d) Solution: (d) S = mid point of Slope (m) of PS = ; The required equation is  2.4 Equation of Parallel and Perpendicular lines to a given Line. (1) Equation of a line which is parallel to is (2) Equation of a line which is perpendicular to is The value of in both cases is obtained with the help of additional information given in the problem. Example: 7 The equation of the line passes through (a, b) and parallel to the line , is (a) (b) (c) (d) Solution: (b) The equation of parallel line to given line is . This line passes through point (a, b).  Hence, required line is . Example: 8 A line passes through (2, 2) and is perpendicular to the line . Its y-intercept is (a) (b) (c) 1 (d) Solution: (d) The equation of a line passing through (2, 2) and perpendicular to is or . Putting in this equation, we obtain . So y-intercept = . Example: 9 The equation of line passing through and perpendicular to is (a) (b) (c) (d) Solution: (a) Equation of a line, perpendicular to is It is passing through . Hence,   . Example: 10 The equation of the line bisecting perpendicularly the segment joining the points (– 4, 6) and (8, 8) is (a) (b) (c) (d) Solution: (a) Equation of the line passing through (–4, 6) and (8, 8) is   ......(i) Now equation of any line to it is = 0 .....(ii) This line passes through the midpoint of (–4, 6) and (8, 8) i.e., (2, 7) From (ii) 12 + 7 +  , Equation of line is 2.5 General equation of a Straight line and its Transformation in Standard forms. General form of equation of a line is , its (1) Slope intercept form: , slope and intercept on y-axis is, (2) Intercept form : , x intercept is = and y intercept is = (3) Normal form : To change the general form of a line into normal form, first take c to right hand side and make it positive, then divide the whole equation by like where , and 2.6 Selection of Co-ordinate of a Point on a Straight line. (1) If the equation of the straight line be , in order to select a point on it, take the x co-ordinate according to your sweet will. Let ; then or ; is a point on the line for any real value of . If is taken then the point will be . Similarly a suitable point can be taken as . (2) If the equation of the line be then a point on it can be taken as where has any real value. In particular (c, 0) is a convenient point on it when . (3) If the equation of the line be then a point on it can be taken as where has any real value. In particular (0, c) is a convenient point on it when . Example: 11 If we reduce to the form , then the value of p is (a) (b) (c) (d) Solution: (d) Given equation is , Dividing both sides by   , . 2.7 Point of Intersection of Two lines. Let =0 and be two non-parallel lines. If be the co-ordinates of their point of intersection, then and Solving these equation, we get Note :  Here lines are not parallel, they have unequal slopes, then .  In solving numerical questions, we should not remember the co-ordinates given above, but we solve the equations directly. 2.8 General equation of Lines through the Intersection of Two given Lines. If equation of two lines and , then the equation of the lines passing through the point of intersection of these lines is or ; Value of is obtained with the help of the additional information given in the problem. Example: 12 Equation of a line passing through the point of intersection of lines , and perpendicular to , then its equation is (a) (b) (c) (d) None of these Solution: (b) The point of intersection of the lines and is The slope of required line = . Hence, Equation of required line is,  . Example: 13 The equation of straight line passing through point of intersection of the straight lines and and having infinite slope is (a) (b) (c) (d) Solution: (c) Required line should be, .....(i)   ......(ii) As the equation (ii) has infinite slope,  Putting in equation (i), We have  2.9 Angle between Two non-parallel Lines. Let  be the angle between the lines and . and intersecting at A. where, = and =   . (1) Angle between two straight lines when their equations are given : The angle  between the lines and is given by, . (i) Condition for the lines to be parallel : If the lines and are parallel then, . (ii) Condition for the lines to be perpendicular : If the lines and are perpendicular then, . (iii) Conditions for two lines to be coincident, parallel, perpendicular and intersecting : Two lines and are, (a) Coincident, if (b) Parallel, if (c) Intersecting, if (d) Perpendicular, if Example: 14 Angle between the lines and is (a) (b) (c) (d) Solution: (b)  . Example: 15 To which of the following types the straight lines represented by and belongs (a) Parallel to each other (b) Perpendicular to each other (c) Inclined at 45° to each other (d) Coincident pair of straight lines Solution: (a) Here, ; . Hence, lines are parallel to each other. 2.10 Equation of Straight line through a given point making a given Angle with a given Line. Since straight line makes an angle with x-axis, then equation of line L is and straight line makes an angle with x-axis, then equation of line is  where Hence, the equation of the straight lines which pass through a given point and make a given angle with given straight line are Example: 16 The equation of the lines which passes through the point (3, – 2) and are inclined at to the line (a) (b) (c) (d) None of these Solution: (a) The equation of lines passing through (3, –2) is .....(i) The slope of the given line is . So, . On solving, we get or Putting the values of m in (i), the required equation is and . Example: 17 In an isosceles triangle ABC, the coordinates of the point B and C on the base BC are respectively (1, 2) and (2, 1). If the equation of the line AB is , then the equation of the line AC is (a) (b) (c) (d) Solution: (b) Slope of BC = ,     . But slope of AB is 2; (Here m is the gradient of the line AC) Equation of the line AC is  or . 2.11 A Line equally inclined with Two lines. Let the two lines with slopes and be equally inclined to a line with slope m then , Note :  Sign of m in both brackets is same. Example: 18 If the lines and are equally inclined to the line , then = (a) (b) (c) (d) Solution: (d) If line are equally inclined to lines with slope and ,then  2.12 Equations of the bisectors of the Angles between two Straight lines. The equation of the bisectors of the angles between the lines and are given by, .....(i) Algorithm to find the bisector of the angle containing the origin : Let the equations of the two lines and . To find the bisector of the angle containing the origin, we proceed as follows: Step I : See whether the constant terms and in the equations of two lines positive or not. If not, then multiply both the sides of the equation by –1 to make the constant term positive. Step II : Now obtain the bisector corresponding to the positive sign i.e., . This is the required bisector of the angle containing the origin. Note :  The bisector of the angle containing the origin means the bisector of the angle between the lines which contains the origin within it. (1) To find the acute and obtuse angle bisectors Let be the angle between one of the lines and one of the bisectors given by (i). Find . If , then this bisector is the bisector of acute angle and the other one is the bisector of the obtuse angle. If > 1, then this bisector is the bisector of obtuse angle and other one is the bisector of the acute angle. (2) Method to find acute angle bisector and obtuse angle bisector (i) Make the constant term positive, if not. (ii) Now determine the sign of the expression . (iii) If , then the bisector corresponding to “+” sign gives the obtuse angle bisector and the bisector corresponding to “–” sign is the bisector of acute angle between the lines. (iv) If , then the bisector corresponding to “+” and “–” sign given the acute and obtuse angle bisectors respectively. Note :  Bisectors are perpendicular to each other.  If , then the origin lies in obtuse angle and if , then the origin lies in acute angle. Example: 19 The equation of the bisectors of the angles between the lines are (a) and (b) and (c) and (d) None of these Solution: (c) The equation of lines are and . The equation of bisectors of the angles between these lines are  Taking +ve sign, we get ; Taking –ve sign, we get . Hence, the equation of bisectors are . Example: 20 The equation of the bisector of the acute angle between the lines and is (a) (b) (c) (d) Solution: (b) Bisector of the angles is given by  .....(i) and ......(ii) Let the angle between the line and (i) is , then Hence is the bisector of the acute angle between the given lines. 2.13 Length of Perpendicular. (1) Distance of a point from a line : The length p of the perpendicular from the point to the line is given by . Note :  Length of perpendicular from origin to the line is .  Length of perpendicular from the point to the line is . (2) Distance between two parallel lines : Let the two parallel lines be and . First Method : The distance between the lines is . Second Method : The distance between the lines is , where (i) if they be on the same side of origin. (ii) if the origin O lies between them. Third method : Find the coordinates of any point on one of the given line, preferably putting or . Then the perpendicular distance of this point from the other line is the required distance between the lines. Note::  Distance between two parallel lines and is  Distance between two non parallel lines is always zero. 2.14 Position of a Point with respect to a Line. Let the given line be and observing point is , then (i) If the same sign is found by putting in equation of line and , then the point is situated on the side of origin. (ii) If the opposite sign is found by putting in equation of line and , then the point is situated opposite side to origin. 2.15 Position of Two points with respect to a Line. Two points and are on the same side or on the opposite side of the straight line according as the values of and are of the same sign or opposite sign. Example: 21 The distance of the point (–2, 3) from the line is (a) (b) (c) (d) Solution: (a) Example: 22 The distance between the lines and is (a) (b) 4 (c) (d) None of these Solution: (c) Given lines and , distance from the origin to both the lines are and  , Clearly both lines are on the same side of the origin. Hence, distance between both the lines are, . Example: 23 If the length of the perpendicular drawn from origin to the line whose intercepts on the axes are a and b be p, then (a) (b) (c) (d) Solution: (d) Equation of line is  Perpendicular distance from origin to given line is    Example: 24 The point on the x-axis whose perpendicular distance from the line is a, is (a) (b) (c) (d) None of these Solution: (a) Let the point be then   Hence the point is Example: 25 The vertex of an equilateral triangle is (2, –1) and the equation of its base is . The length of its sides is (a) (b) (c) (d) None of these Solution: (b)    2.16 Concurrent Lines. Three or more lines are said to be concurrent lines if they meet at a point. First method : Find the point of intersection of any two lines by solving them simultaneously. If the point satisfies the third equation also, then the given lines are concurrent. Second method : The three lines , and are concurrent if, Third method : The condition for the lines , and to be concurrent is that three constants a, b, c (not all zero at the same time) can be obtained such that . Example: 26 If the lines , and be concurrent, then (a) (b) (c) (d) None of these Solution: (c) Here the given lines are, , , The lines will be concurrent, iff  Example: 27 If the lines and are concurrent, then b equals (a) 1 (b) 3 (c) 6 (d) 0 Solution: (c) If these lines are concurrent then the intersection point of the lines and , is (–2, 3), which lies on the third line. Hence,     Example: 28 The straight lines where , will be concurrent, if point is (a) (4, 3) (b) (c) (d) None of these Solution: (b) The set of lines is , where Eliminating c, we get  They pass through the intersection of the lines and i.e., i.e., 2.17 Reflection on the Surface. Here IP = Incident Ray PN = Normal to the surface PR = Reflected Ray Then, Angle of incidence = Angle of reflection 2.18 Image of a Point in Different cases. (1) The image of a point with respect to the line mirror : The image of with respect to the line mirror be B (h, k) is given by, (2) The image of a point with respect to x-axis : Let be any point and its image after reflection in the x-axis, then = x ( is the mid point of P and ) = – y (3) The image of a point with respect to y-axis : Let be any point and its image after reflection in the y-axis then ( is the mid point of P and ) (4) The image of a point with respect to the origin : Let be any point and be its image after reflection through the origin, then ( is the mid point of P and ) (5) The image of a point with respect to the line y = x : Let be any point and be its image after reflection in the line , then ( is the mid point of P and ) (6) The image of a point with respect to the line y = x tan  : Let be any point and be its image after reflection in the line then ( is the mid point of P and ) Example: 29 The reflection of the point (4, –13) in the line is (a) (–1, –14) (b) (3, 4) (c) (1, 2) (d) (–4, 13) Solution: (a) Let be the reflection of in the line . Then the point lies on .  .....(i) Also PQ is perpendicular to . Therefore  .....(ii) Solving (i) and (ii), we get . Example: 30 The image of a point in the line , is (a) (–1, –4) (b) (–3, –8) (c) (1, –4) (d) (3, 8) Solution: (a) Equation of the line passing through (3, 8) and perpendicular to is . The intersection point of both the lines is (1, 2). Now let the image of be . The point (1, 2) will be the midpoint of .  and  . Hence the image is (–1, –4). 2.19 Some Important Results. (1) Area of the triangle formed by the lines , is . (2) Area of the triangle made by the line with the co-ordinate axes is . (3) Area of the rhombus formed by the lines is (4) Area of the parallelogram formed by the lines ; , and is . (5) The foot of the perpendicular from to the line is given by . Hence, the coordinates of the foot of perpendicular is (6) Area of parallelogram , where and are the distances between parallel sides and is the angle between two adjacent sides. (7) The equation of a line whose mid-point is in between the axes is (8) The equation of a straight line which makes a triangle with the axes of centroid is . Example: 31 The coordinates of the foot of perpendicular drawn from (2, 4) to the line is (a) (b) (c) (d) Solution: (b) Applying the formula, the required co-ordinates is = Example: 32 The area enclosed within the curve is (a) (b) 1 (c) (d) 2 Solution: (d) The given lines are i.e., , , and . These lines form a quadrilateral whose vertices are , , and . Obviously ABCD is a square. Length of each side of this square is . Hence, area of square is sq. units. Example: 33 If and are both in G.P. with the same common ratio, then the point , and (a) Lie on a straight line (b) Lie on an ellipse (c) Lie on a circle (d) Are vertices of a triangle Solution: (a) Taking co-ordinates as , (x, y) and . Above co-ordinates satisfy the relation , the three points lie on a straight line. Example: 34 A square of side a lies above the x-axis and has one vertex at the origin. The side passing through the origin makes an angle with the positive direction of x-axis. The equation of its diagonal not passing through the origin is (a) (b) (c) (d) Solution: (b) Co-ordinates of ; Equation of to OB ; slope of Equation of CA,  . Example: 35 The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle with vertices (0, 0), (0, 21) and (21, 0) is (a) 133 (b) 190 (c) 233 (d) 105 Solution: (b) The number of integral solution to the equation i.e., Number of integral co-ordinates = .