Straight lines MODULE - II Coordinate Geometry STRAIGHT LINES

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1 Straight lines 0 STRIGHT LINES We have read about lines, angles and rectilinear figures in geometry. Recall that a line is the join of two points in a plane continuing endlessly in both directions. We have also seen that graphs of linear equations, which came out to be straight lines Interestingly, the reverse problem of the above is finding the equations of straight lines, under different conditions, in a plane. The analytical geometry, more commonly called coordinate geomatry, comes to our help in this regard. In this lesson. We shall find equations of a straight line in different forms and try to solve problems based on those. OJECTIVES fter studying this lesson, you will be able to : derive equations of a line parallel to either of the coordinate axes; derive equations in different forms (slope-intercept, point-slope, two point, intercept, parametric and perpendicular) of a line; find the equation of a line in the above forms under given conditions; state that the general equation of first degree represents a line; express the general equation of a line into (i) slope-intercept form (ii) intercept form and (iii) perpendicular form; derive the formula for the angle between two lines with given slopes; find the angle between two lines with given slopes; derive the conditions for parallelism and perpendicularity of two lines; determine whether two given lines are parallel or perpendicular; derive an expression for finding the distance of a given point from a given line; calculate the distance of a given point from a given line; MTHEMTICS 339

2 Straight lines derive the equation of a line passing through a given point and parallel/perpendicular to a given line; write the equation of a line passing through a given point and: (i) parallel or perpendicular to a given line (ii) with given x-intercept or y-intercept (iii) passing through the point of intersection of two lines; and prove various geometrical results using coordinate geometry. EXPECTED CKGROUND KNOWLEDGE Congruence and similarity of traingles 0. STRINGHT LINE PRLLEL TO N XIS If you stand in a room with your arms stretched, we can have a line drawn on the floor parallel to one side. nother line perpendicular to this line can be drawn intersecting the first line between your legs. In this situation the part of the line in front of you and going behind you is they-axis and the one being parallel to your arms is the x-axis. The direction part of the y-axis in front of you is positive and behind you is negative. The direction of the part x-axis to your right is positive and to that to your left is negative. Now, let the side facing you be at b units away from you, then the equation of this edge will be y = b (parallel to x-axis) where b is equal in absolute value to the distance from thex-axis to the opposite side. If b > 0, then the line lies in front of you, i.e., above thex-axis. If b < 0, then the line lies behind you, i.e., below thex-axis. If b = 0, then the line passes through you and is the x-axis itself. gain, let the side of the right of you is atc units apart from you, then the equation of this line will be x = c (parallel to y - axis) where c is equal in absolute value, to the distance from they-axis on your right. If c > 0, then the line lies on the right of you, i.e., to the right ofy-axis. If c < 0, then the line lies on the left of you, i.e., to the left ofy-axis If c = 0, then the line passes through you and is the y-axis. Example 0. Find the equation of the lines passing through (, 3) and is (i) parallel tox-axis. (ii) parallel to y-axis. 340 MTHEMTICS

3 Straight lines Solution : (i) The equation of any line parallel to x-axis is y = b Since it passes through (, 3), henceb = 3 The required equation of the line is y = 3 (ii) The equation of any line parallel to y-axis is x = c Since it passes through (, 3), hencec = The required equation of the line is x =. Example 0. Find the equation of the line passing through (, 3) and (i) parallel tox-axis (ii) parallel to y-axis Solution : (i) The equation of any line parallel to x-axis is y = b Since it passes through (, 3), hence 3 =b The required equation of the line is y = 3 (i) The equation of any line parallel to y-axis is x = c Since it passes through (, 3), hence =c The required cquation of the line is x = CHECK YOUR PROGRESS 0.. If we fold and press the paper then what will the crease look like. Find the slope of a line which makes an angle of (a) 45 with the positive direction of x-axis. (b) 45 with the positive direction of y-axis. (c) 45 with the negative direction of x-axis. 3. Find the slope of a line joining the points (, 3) and (3, 4). 4. Determine x so that the slope of the line through the points (,5) and (7,x) is Find the equation of the line passing through ( 3, 4) and (a) parallel to x-axis. (b) parallel to y-axis. 6. Find the equation of a line passing through (5, 3) and perpendicular to x-axis. 7. Find the equation of the line passing through ( 3, 7) and perpendicular toy-axis. MTHEMTICS 34

4 Straight lines 0. DERIVTION OF THE EQUTION OF STRIGHT LINE IN VRIOUS STNDRD FORMS So far we have studied about the inclination, slope of a line and the lines parallel to the axes. Now the questions is, can we find a relationship betweenx and y, where (x, y) is any arbitrary point on the line? The relationship between x and y which is satisfied by the co-ordinates of arbitrary point on the line is called the equation of a straight line. The equation of the line can be found in various forms under the given conditions, such as (a) (b) (c) (d) (e) (f) () When we are given the slope of the line and its intercept ony-axis. When we are given the slope of the line and it passes through a given point. When the line passes through two given points. When we are given the intercepts on the axes by the line. When we are given the length of perpendicular from origin on the line and the angle which the perpendicualr makes with the positive direction ofx-axis. When the line passes through a given point making an angleα with the positive direction of x-axis. (Parametric form). We will discuss all the above cases one by one and try to find the equation of line in its standard forms. SLOPE-INTECEPT FORM Let be a straight line making an angleθ with x-axis and cutting off an interceptod = c from OY. s the line makes intercept OD = c on y-axis, it is called y-intercept. Let intersect OX' at T. Take any point P(x, y) on. Draw PM OX. The OM = x, MP = y. Draw DN MP. From the right-angled triangle DNP, we have tan θ = NP DN MP MN = OM D Y θ P N = y OD OM = y c X' T O x M X x Fig. 0. Y' θ c y 34 MTHEMTICS

5 Straight lines y = x tan θ + c tan θ = m (slope) y = mx + c Since, this equation is true for every point on, and clearly for no other point in the plane, hence it represents the equation of the line. Note : () When c = 0 and m 0 the line passes through the origin and its equation is y = mx () When c = 0 and m = 0 the line coincides with x axis and its equation is of the form y = o (3) Whenc 0 and m = 0 the line is parallel to x-axis and its equation is of the form y = c Example 0.3 Find the equation of a line with slope 4 and y-intercept 0. Solution : Putting m = 4 and c = 0 in the slope intercept form of the equation, we get y = 4 x This is the desired equation of the line. Example 0.4 Determine the slope and the y-intercept of the line whose equation is 8x + 3y = 5. Solution : The given equation of the line is 8x + 3y = 5 or, y = 8 3 x Y Comparing this equation with the equation y = mx + c (Slope intercept form) we get m = 8 3 and c 5 = 3 X' 0 0 O X Therefore, slope of the line is 8 3 and its y-intercept is 5 3. Fig. 0. Y' Example 0.5 Find the equation of the line cutting off an intercept of length from the negative direction of the axis ofy and making an angle of 0 0 with the positive directionx-axis Solution : From the slope intercept form of the line y = x tan 0 + ( ) MTHEMTICS 343

6 = 3 x or, y + 3 x + = 0 Straight lines Here m = tan 0, and c =, because the intercept is cut on the negative side of y -axis. (b) POINT-SLOPE FORM Here we will find the equation of a line passing through a given point (x, y ) and having the slope m. Let P(x, y) be any point other than on given the line. Slope (tan) of the line joining (x, y ) and P (x, y) is given by m = tan = y x y x O ( x, y ) P( x y) θ θ x x M y y The slope of the line P is given to be m. m = y y x x Fig. 0.3 The equation of the required line is y y = m (x x ) Note : Since, the slope m is undefined for lines parallel to y-axis, the point-slope form of the equation will not give the equation of a line though (x, y ) parallel to y-axis. However, this presents no difficulty, since for any such line the abscissa of any point on the line is x. Therefore, the equation of such a line is x = x. Example 0.6 Determine the equation of the line passing through the point (, ) and having slope 3. Solution : Putting x =, y = and m = 3 in the equation of the point-slope form of the line we get y ( ) = 3 ( x ) y + = 3 ( x ) 344 MTHEMTICS

7 Straight lines y = 3 x 7 3 which is the required equation of the line. (c) TWO POINT FORM Let (x, y ) and (x, y ) be two given distinct points. Slope of the line passing through these points is given by m = y x y x (x x ) From the equation of line in point slope form, we get y y y y = () x x x x which is the required equation of the line in two-point form. Example 0.7 Find the equation of the line passing through (3, 7) and (, 5). Solution : The equation of a line passing through two points (x, y ) and (x, y ) is given by y y = y x y x (x x ) (i) Since x = 3, y = 7 and x =, and y = 5, equation (i) becomes, y + 7 = (x 3) or, y + 7 = 5 (x 3) or, x + 5 y + 9 = 0 (d) INTERCEPT FORM We want to find the equation of a line which cuts off given intercepts on both the co-ordinate axes. P b Y ( 0,b) Let PQ be a line meeting x-axis in and y-axis in. Let O = a, O = b. Then the co-ordinates of and are (a,0) and (0, b,) respectively. X' O Y' a (a,)0 Q Fig. 0.4 X MTHEMTICS 345

8 The equation of the line joining and is y 0 = b 0 0 a (x a) Straight lines or, y = b a (x a) or, y b = x a + or, x a + y b = This is the required equation of the line having interceptsa and b on the axes. Example 0.8 Find the equation of a line which cuts off intercepts 5 and 3 on x and y axes respectively. Solution : The intercepts are 5 and 3 on x and y axes respectively. i.e., a = 5, b = 3 The required equation of the line is x y + = x 5y 5 = 0 Example 0.9 Find the equation of a line which passes through the point (3, 4) and makes intercepts on the axes equl in magnitude but opposite in sign. Solution : Let the x-intercept and y-intercept be a and a respectively The equation of the line is x y + = a a x y = a (i) Since (i) passes through (3, 4) 3 4 = a or or, a = Thus, the required equation of the line is x y = or x y + = MTHEMTICS

9 Straight lines Example 0.0 Determine the equation of the line through the point (,) and parallel to x - axis. Solution : Since the line is parallel to x-axis its slope ia zero. Therefore from the point slope form of the equation, we get y = 0 [ x ( )] y = 0 which is the required equation of the given line Example 0. Find the intercepts made by the line 3x y + = 0 on the coordinate axes Solution : Equation of the given line is 3x y =. Dividing by, we get x y + = 4 6 Comparing it with the standard equation of the line in intercept form, we finda = 4 and b = 6. Hence the intercepts on the x-axis and y-axis repectively are 4. and 6. Example 0. The segment of a line, intercepted between the coordinate axes is bisected at the point (x, y ). Find the equation of the line Solution : Let P(x, y ) be the middle point or the segment CD of the line intercepted between the axes. Draw PM OX Y OM = x and MP = y OC = x and OD = y Now, from the intercept form of the line D y P ( x, y ) x x + y y = X' O x M C X or, x x + y y = Fig. 0.5 Y' which is the required equation of the line. MTHEMTICS 347

10 Straight lines (e) PERPENDICULR FORM (NORML FORM) We now derive the equation of a line when p be the length of perpendicular from the origin on the line and, the angle which this perpendicular makes with the positive direction of x-a x is are given. Y Y b P p b 80- p P( x, y ) X' a O X X' O Y' a (i) X Y' Fig. 0.6 (ii) (i) Fig. 0.6 Let be the given line cutting off interceptsa and b on x axis and y axis respectively. Let OP be perpendicular from origin O on and PO = (See Fig. 0.6 (i)) p a = cos a = p sec and, or, (ii) p b = sin b = p cosec The equation of line is x y + = p sec p cos ec x cos + y cosec = p p a = cos (80 ) = cos a = p sec similary, b = p cosec The equation of the line is x y a + b = x cos + y sin = p [From Fig. 0.6 (ii)] 348 MTHEMTICS

11 Straight lines Note :. p is the length of perpendicular from the origin on the line and in always taken to be positive.. is the angle between positive direction of x-axis and the line perpendicular from the origin to the given line. Example 0.3 Determine the equation of the line with = 35 and perpendicular distance p = from the origin. Solution : From the standard equation of the line in normal form have x cos 35 + y sin 35 = or, x + y = or, x + y = 0 or, x y + = 0 which is the required equation of the straight line. Example 0.4 Find the equation of the line whose perpendicular distance from the origin is 6 units and the perpendicular from the origin to line makes an angle of 30 with the positive direction of x-axis. Solution : Here = 30, p = 6 The equation of the line is x cos 30 + y sin 30 = 6 F HG 3 or, x y I KJ F + H G I K J = 6 P(x,y) or, 3 x + y = r (f) PRMETRIC FORM We now want to find the equation of a line through a given point Q (x, y ) which makes an angle with the positive direction of x axis in the form Q( x y) y N y x x y y = = r X' O x L M X cos sin x Fig. 0.7 MTHEMTICS 349

12 where r is the distance of any point P (x, y) on the line from Q (x, y ). Straight lines Let be a line passing through a given point Q (x, y ) making an angle with the positive direction of x-axis. Let P (x, y) be any point on the line such that QP = r (say) Draw QL, PM perpendiculars to x-axis and draw QN PM. From right angled PNQ cos = QN PN, sin QP = QP ut QN = LM = OM OL = x x and PN = PM MN = PM QL = y y cos = x x, sin = y y r r or, x x y y = = r, which is the required equation of the line cos sin Note :. The co-ordinates of any point on this line can be written as (x + r cos, y + r sin ). Clearly coordinates of the point depend on the value of r. This variable r is called parameter.. The equation x = x + r cos, y = y + r sin are called the parametric equations of the line. 3. The value of 'r' is positive for all points lying on one side of the given point and negative for all points lying on the other side of the given point. Example 0.5 Find the equation of a line passing through (, ) and making an angle of 30 with the positive direction of x-axis, in the parametric form. lso find the coordinates of a point P on it at a distance of units from the point. Solution : Here x =, y = and = 30 The equation of the line is x + y = + cos 30 sin 30 or, x + y + = MTHEMTICS

13 Straight lines Since P =, r = x + y + = = 3 x = 3, y = Thus the coordinates of the point P are 3, Example 0.6 Find the distance of the point (, ) from the line x 3y + 9 = 0 measured along a line making an angle of 45 withx axis. Solution : The equation of any line through (, ) making an angle of 45 withx-axis is d i x y r cos 45 = sin 45 = x y = = r F HG r ny point on it is + r, + It lies on the given line x 3y + 9 = 0 F r I r if HG K J I + HG K J + = r or, = 0 or, r = 5 Hence the required distance is 5. F I K J MTHEMTICS 35

14 CHECK YOUR PROGRESS 0.. (a) Find the equation of a line with slope andy intercept is equal to. Straight lines (b) Determine the slope and the intercepts made by the line on the axes whose equation is 4x + 3y = 6.. Find the equation of the line cutting off an interecept on negative direction of axis of 3 y and inclined at 0 to the positive direction ofx-axis. 3. Find the slope and y-intercept of the line whose equation is 3x 6y =. 4. Determine the equation of the line passing through the point ( 7, 4) and having the slope Determine the equation of the line passing through the point (, ) which makes equal angles with the two axes. 6. Find the equation of the line passing through (, 3) and parallel to the line joining the points (, ) and (6, 4). 7. (a) Determine the equation of the line through (3, 4) and ( 4, 3). (b) Find the equation of the diagonals of the rectangle CD whose vertices are (3, ), (, 8), C (8, ) and D (0, 6). 8. Find the equation of the medians of a triangle whose vertices are (, 0), (0, ) and (4, 6). 9. Find the equation of the line which cuts off intercepts of length 3 units and units on x-axis and y-axis respectively. 0. Find the equation of a line such that the segment between the coordinate axes has its mid point at the point (, 3). Find the equation of a line which passes through the point (3, ) and cuts off positive intercepts on x and y axes in the ratio of 4 : 3.. Determine the equation of the line whose perpendicular from the origin is of length units and makes an angle of 45 with the positwe direction ofx-axis. 3. If p is the length of the perpendicular segment from the origin, on the line whose intercept on the axes are a and b, then show that = +. p a b 4. Find the equation of a line passing through (, ) and making an angle 45 with the positive direction of x-axis in parametric form. lso find the coordinates of a pointp on it at a distance of unit from the point. 35 MTHEMTICS

15 Straight lines 0. 3 GENERL EQUTION OF FIRST DEGREE You know that a linear equation in two variables x and y is given by x + y + C = 0...() In order to understand its graphical representation, we need to take the following thres cases. Case-: (When both and are equal to zero) In this case C is automaticaly zero and the equation does not exist. Case-: (When = 0 and 0) In this case the equation () becomes y + C = 0. or y = C and is satisfied by all points lying on a line which is parallel to x-axis and the y-coordinate of every point on the line is C. Hence this is the equation of a straight line. The case where = 0 and 0 can be treated similarly. Case-3: (When 0 and 0) We can solve the equation () fory and obtain. y = x C Clearly, this represents a straight line with slope and y intercept equal to C CONVERSION OF GENERL EQUTION OF LINE INTO VRIOUS FORMS If we are given the general equation of a line, in the form x + y + C = 0, we will see how this can be converted into various forms studied before CONVERSION INTO SLOPE-INTERCEPT FORM We are given a first degree equation in x and y as x + y + C = O re you able to find slope and y-intercept? Yes, indeed, if we are able to put the general equation in slope-intercept form. For this purpose, let us re-arrange the given equation as. x + y + C = 0 as y = x C or y = x C (Provided 0) which is the required form. Hence, the slope =, y intercept = C. MTHEMTICS 353

16 Straight lines Example 0.7 Reduce the equation x + 7y 4 = 0 to the slope intercept form. Solution : The given equation is x + 7y 4 = 0 or 7y = x + 4 or y = 4 x which is the converted form of the given equation in slope-intercept form Example 0.8 Find the slope and y intercept of the line x + 4y 3 = 0. Solution : The given equation is x + 4y 3 = 0 or 4y = x + 3 or y = 4 3 x + 4 Comparing it with slope-intercept form, we have slope = 4, y intercept = CONVERSION INTO INTERCEPT FORM Suppose the given first degree equation in x and y is x + y + C = 0....(i) In order to convert (i) in intercept form, we re arrange it as x + y = C or x y + = C C or x C + y C = ( ) ( ) (Provided 0 and 0) which is the requied converted form. It may be noted that intercept on x axis = C and intercept on y axis = C 354 MTHEMTICS

17 Straight lines Example 0.9 Reduce 3x + 5 y = 7 into the intercept form and find its intercepts on the axes. Solution : The given equation is or, 3x + 5y = x + y = 7 or, x y + = The x intercept = 7 3 and, y intercept = 7 5 Example 0.0 Find x and y-intercepts for the line 3x y = 5. Solution : The given equation is 3x y = 5 or, or, 3 5 x y = 5 x y + = Thus, the required x intercept = 5 3 and y-intercept = CONVERSION INTO PERPENDICULR FORM Let the general first degree equation in x and y be x + y + C = 0... (i) We will convert this general equation in perpendicular form. For this purpose let us re-write the given equation (i) as x + y = C Multiplying both sides of the above equation byλ, we have λx + λy = λc Let us choose λ such that (λ) + (λ) =... (ii) MTHEMTICS 355

18 Straight lines or λ = ( + ) (Taking positive sign) Substituting this value ofλ in (ii), we have x y + ( + ) ( + ) = C ( + )... (iii) This is required conversion of (i) in perpendicular form. Two cases arise according asc is negative or positive. (i) (ii) If C < 0, the equation (ii) is the required form. If C > 0, the R. H. S. of the equation of (iii) is negative. We shall multiply both sides of the equation of (iii) by. The required form will be x y C = ( + ) ( + ) ( + ) Thus, length of perpendicular from the origin = C ( + ) Inclination of the perpendicular with the positwe direction of x-ax is is given by cos θ = F H G + or sin θ = ( + ) I K J where the upper sign is taken for C > 0 and the lower sign for C < 0. If C = 0, the line passes through the origin and there is no perpendicular from the origin on the line. With the help of the above three cases, we are able to say that "The general equation of first degree in x and y always represents a straight line provided and are not both zero simultaneously." Is the converse of the above statement true? The converse of the above statement is that every straight line can be expressed as a general equation of first degree in x and y. 356 MTHEMTICS

19 Straight lines In this lesson we have studied about the various forms of equation of straight line. For example, let us take some of them as y = mx + c, x y + = and x cos α + y sin α = p. Obviously, all a b are linear equations in x and y. We can re-arrange them as y mx c = 0, bx + ay ab = 0 and x cos α + y sin α p = 0 respectively. Clearly, these equations are nothing but a different arrangement of general equation of first degree inx and y. Thus, we have established that "Every straight line can be expressed as a general equation of first degree in x and y". Example 0. Reduce the equation x + 3 y + 7 = 0 into perpendicular form. Solution : The equation of given line is x + 3y + 7 = 0... (i) Comparing (i) with general equation of straight line, we have = and = 3 + = Dividing equation (i) by, we have x 3 y = 0 F I HG K J F + H G I K J = 3 7 or x y or x cos y sin = (cos θ and sinθ being both negative in the third quadrant, value ofθ will lie in the third quadrant). This is the representation of the given line in perpendicular form. Example 0. Find the perpendicular distance from the origin on the line 3 x y + = 0. lso, find the inclination of the perpendicular from the origin. Solution : The given equation is 3 x y + = 0 Dividing both sides by ( 3) + ( ) or, we have MTHEMTICS 357

20 3 x y + = 0 Straight lines or, 3 x y = Multiplying both sides by, we have 3 x + y = or, x cos y sin = (cosθ is ve in second quadrant and sin θ is +ve in second 6 6 quadrant, so value of θ lies in the second quadrant). Thus, inclination of the perpendicular from the origin is 50 and its length is equal to. Example 0.3 Find the equation of a line which passess through the point (3,) and bisects the portion of the line 3x + 4y = intercepted between coordinate axes. Solution : First we find the intercepts on coordinate axes cut off by the line whose equation is from. 3x + 4y = or 3x 4y + = or x y + = 4 3 Hence, intercepts on x-axis and y-axis are 4 and 3 respectively. Thus, the coordinates of the points where the line meets the coordinate axes are (4, 0) and (0, 3). F I HG K J. Mid point of is, 3 Hence the equation of the line through (3, ) F HG and, 3 I K J is y = ( x ) (0,3) O y D Fig , C (3,) (4,0) x 358 MTHEMTICS

21 Straight lines or y = ( x 3) or (y ) + (x 3) = 0 or y + x 3 = 0 or x + y 5 = 0 Example 0.4 Prove that the line through (8, 7) and (6, 9) cuts off equal intercepts on coordinate axes. Solution : The equation of the line passing through (8, 7) and (6, 9) is 9 7 y 7 = ( x 8) 6 8 or y 7 = (x 8) or x + y = 5 x y or = Hence, intercepts on both axes are 5 each. Example 0.5 Find the ratio in which the line joining ( 5, ) and (, 3) divides the join of (3, 4) and (7, 8). Solution : The equation of the line joining C ( 5, ) and D (, 3) is 3 y = ( x + 5 ) + 5 or 4 y ( = 5) x + 6 or 3y 3 = x 0 or x + 3y + 7 = 0... (i) Let line (i) divide the join of (3, 4) and (7, 8) at the point P. If the required ratio is λ : in which line (i) divides the join of (3, 4) and (7, 8), then the coordinates of P are F HG , + + I KJ C( 5,) P(x,y) o Y D (3,4) (, 3) Fig. 0.9 (7,8) X MTHEMTICS 359

22 L e t XE Since P lies on the line (i), we have F HG I + KJ F HG = 0 + KJ 4λ λ + + 7λ + 7 = 0 I Straight lines 45λ + 5 = 0 λ = 5 9 Hence, the line joining ( 5, ) and (, 3) divides the join of (3, 4) and (7, 8) externally in the ratio 5 : 9. CHECK YOUR PROGRESS 0.3. Under what condition, the general equation x + y + C = 0 of first degree in x and y represents a line?. Reduce the equation x + 5y + 3 = 0 to the slope intercept form. 3. Find the x and y intercepts for the following lines : (a) y = mx + c (b) 3y = 3x + 8(c) 3x y + = 0 4. Find the length of the line segment intercepted by the straight line 3x y + = 0 between the two axes. 5. Reduce the equation x cos + y sin = p to the intercept form of the equation and also find the intercepts on the axes. 6. Reduce the following equations into normal form. (a) 3x 4y + 0 = 0 (b) 3x 4y = 0 7. Which of the lines x y + 3 = 0 and x 4y 7 = 0 is nearer from the origin? We will here develop a formula for finding angle between two given lines 0.4 NGLE ETWEEN TWO LINES We will now try to find out the angle between the arms of the divider or clock when equation of the arms are known. Two methods are discussed below When two lines with slopes m and m are given Let and CD be two straight lines whose equations are y = m x + c and y = m x + c respectively. Let P be the point of intersection of and CD. = θ and XFD = θ. Then 360 MTHEMTICS

23 Straight lines tan θ = m and tan θ = m Let θ be the angle EPF between and CD. Now θ = θ + θ or θ = θ θ tan θ = tan (θ θ ) tan tan or tan θ = + tan tan x' C F y O P θ θ θ E O D x Hence, tan θ = m m + m m which gives tangent of the angleθ between two given lines in terms of their slopes. From this result we can find the angle between any two given lines when their slopes are given. y' Fig. 0.0 Note : (i) m m If + m m > 0, then angle between lines is acute. (ii) If m m + m m < 0, then angle between lines is obtuse. (iii) m m If + m m = 0, i.e., m = m, then lines are either coincident or parallel. i.e, Two lines are parallel if they differ only by a constant term. (iv) If the denominator of m m + m m is zero. i.e, m m =, then lines are perpendicular. Example 0.6 Find the angle between the lines 7x y = and 6x y =. Solution : Let θ be the angle betwen the lines 7x y = and 6x y =. The equations of these lines can be written as y = 7 x and y = 6x. Here, m = 7, m = 6. where m and m are respective slopes of the given lines. tan θ = 7 6 = MTHEMTICS 36

24 The angle θ between the given lines is given by tan θ = 43 Straight lines Example 0.7 Find the angle between the lines x + y + = 0 and x y = 0. Solution : Putting the equations of the given lines in slope intercept form, we get y = x and y = x m = and m = where m, and m are respectively slopes of the given lines. tan θ = + ( ) ( ) = 3 The angle θ between the given lines is given by tan θ = 3 Example 0.8 re the straight lines y = 3x 5 and 3x + 4 = y, parallel or perpendicular? Solution : For the first line y = 3x 5, the slope m = 3 and for the second line 3x + 4 = y, the slope m = 3. Since the slopes of both the lines are same for both the lines, hence the lines are parallel. Example 0.9 re the straight lines y = 3x and y = 3 x, parallel or perpendicular? Solution : The slope m of the line y = 3x is 3 and slope m of the second line y = x 3 is 3. Since, m m =, the lines are perpendicular When the Equation of two lines are in general form Let the general equations of two given straight lines be given by x + y + C = 0... (i) and x + y + C = 0...(ii) Equations (i) and (ii) can be written into their slope intercept from as y = x C and y x C = Let m and m be their respective slopes and using the formula, we get. i.e., tan = + Thus, by using this formula we can calculate the angle between two lines when they are in general form. 36 MTHEMTICS

25 Straight lines Note : (i) If = 0 or = 0 (ii) i.e. = 0, then the given lines are parallel. If + = 0, then the given lines are perpendicular. Example 0.30 re the straight lines x 7y + = 0 and x 7 y + 6 = 0, parallel? Solution : Here =, = 7; = and = 7 Since = 7 7 = 0 Example 0.3 Determine whether the straight lines x + 5y 8 = 0 and 5x y 3 = 0 are parallel or perpendicular? Here =, = 5 = 5 and = Solution : Here + = () (5) + (5) ( ) = 0. Hence the given lines are perpendicular. CHECK YOUR PROGRESS 0.4. What is the condition for given lines y = m x + c and x + y + C = 0 (a) to be parallel? (b) to be perpendicular?. Find the angle between the lines 4x + y = 3 and x + y = (a) Write the condition that the two lines x + y + C = 0 and x + y + C = 0 are (i) parallel. (ii) perpendicular. (b) re the straight lines x 3y = 7 and x 6y 6 = 0, parallel? (c) re the straight lines x = y + and x = y + perpendicular? MTHEMTICS 363

26 Straight lines 0.5 DISTNCE OF GIVEN POINT FROM GIVEN LINE In this section, we shall discuss the concept of finding the distance of a given point from a given line or lines. Let P(x, y ) be the given point and l be the line x + y + C = 0. Let the line l intersect x axis and y axis R and Q respectively. Draw PM l and let PM = d. Let the coordinates of M be (x, y ) b g b g } d = { x x + y y...(i) M lies on l x + y + C = 0 or C = (x + y )...(ii) C The coordinates of R and Q are, 0 F I HG K J F and 0, C I HG K J respectively. The slope of QR = C 0 + = C 0 and, y y the slope of PM = x x s PM QR y y =. x x y or y x y x =...(iii) Q M d P( x y) x' O R x y' Fig MTHEMTICS

27 Straight lines From (iii) x x y y = = {b x xg + by yg } c + h...(iv) (Using properties of Ratio and Proportion) b g b g lso x x y y x x + y y = = + From (iv) and (v), we get...(v) {b x xg by yg } c + h = b g b g x x + y y + or d + + x + y ( x + y + [Using (i)] or x + y + C c + [Using (ii)] h Since the distance is always positive, we can write d = x + y + C c + h Note : The perpendicular distance of the origin (0, 0) from x + y + C = 0 is ( 0) + ( 0) + C C = c + h ( + ) Example 0.3 Find the points on the x-axis whose perpendicular distance from the straight line x a y + = is a. b Solution : Let (x, 0) be any point on x-axis. Equation of the given line is bx + ay ab = 0. The perpendicular distance of the point (x, 0) from the given line is MTHEMTICS 365

28 bx + a. 0 ab a = ± a + b x c RST c h} a = b b ± a + b Thus, the point on x-axis is h F HG e j, a b b ± a + b 0 I K J Straight lines CHECK YOUR PROGRESS 0.5. Find the perpendicular distance of the point (, 3) from 3x + y + 4 = 0.. Find the points on the axis of y whose perpendicular distance from the straight line x a y + = is b. b 3. Find the points on the axis of y whose perpendicular distance from the straight line 4x + 3y = is Find the perpendicular distance of the origin from 3x + 7y + 4 = 0 5. What are the points on the axis of x whose perpendicular distance from the straight line x y + = is 3? EQUTION OF PRLLEL (OR PERPENDICULR) LINES Till now, we have developed methods to find out whether the given lines are prallel or perpendicular. In this section, we shall try to find, the equation of a line which is parallel or perpendicular to a given line EQUTION OF STRIGHT LINE PRLLEL TO THE GIVEN LINE x + y + c = 0 Let x + y + C = 0...(i) be any line parallel to the given line x + y + C = 0 The condition for parallelism of (i) and (ii) is = = K (say) = K, = K...(ii) 366 MTHEMTICS

29 Straight lines with these values of and, (i) gives K x + K y + C = 0 or x + y + C K = 0 or x + y + K = 0, where K = C K... (iii) This is a line parallel to the given line. From equations (ii) and (iii) we observe that (i) coefficients of x and y are same (ii) constants are different, and are to evaluated from given conditions. Example 0.33 Find equation of the straight line, which passes through the point (, ) and which is parallel to the straight line x + 3y + 6 = 0. Solution : Equation of any straight line parallel to the given equation can be written if we put (i) the coefficients of x and y as same as in the given equation. (ii) constant to be different from the given equation, which is to be evaluated under given condition. Thus, the required equation of the line will be x + 3y + K = 0 for some constant K Since it passes through the point (, ) hence K = 0 or K = 8 Required equation of the line is x + 3y = STRIGHT LINE PERPENDICULR TO THE GIVEN LINE x + y + C = 0 Let x + y + C = 0 be any line perpendicular to the given line x + y + C = 0 Condition for perpendicularity of lines (i) and (ii) is + = 0... (i)... (ii) = = K (say) MTHEMTICS 367

30 = K and = K With these values of and, (i) gives x y + C K = 0 = 0 Straight lines or x y + K = 0 where K = C K... (iii) Hence, the line (iii) is perpendicular to the given line (ii) We observe that in order to get a line perpendicular to the given line we have to follow the following procedure : (i) Interchange the coefficients of x and y (ii) Change the sign of one of them. (iii) Change the Constant term to a new constant K (say), and evaluate it from given condition. Example 0.34 Find the equation of the line which passes through the point (, ) and is perpendicular to the line x + 3y + 6 = 0. Solution : Following the procedure given above, we get the equation of line perpendicular to the given equation as 3x y + K = 0...(i) (i) passes through the point (, ), hence 3 + K = 0 or K = Required equation of the straight line is 3x y + = 0. Example 0.35 Find the equation of the line which passes through the point (x, y ) and is perpendicular to the straight line y y = a (x + x ). Solution : The given straight line is yy ax ax = 0...(i) ny straight line perpendicular to (i) is ay + xy + C = 0 This passes through the point (x, y ) ay + x y + C = 0 C = ay x y Required equation of the spraight line is a (y y ) + y (x x ) = MTHEMTICS

31 Straight lines CHECK YOUR PROGRESS 0.6. Find the equation of the straight line which passes through the point (0, ) and is parallel to the straight line 3x + y =.. Find the equation of the straight line which passes through the point (, 0) and is parallel to the straight line y = x Find the equation of the straight line which passes through the point (0, 3) and is perpendicular to the straight line x + y + = Find the equation of the line which passes through the point (0, 0) and is perpendicular to the straight line x + y = Find the equation of the straight line which passes through the point (, 3) and is perpendicular to the given straight line a (x + ) + 3y = Find the equation of the line which has x - intercept 8 and is perpendicular to the line 3x + 4y 7 = Find the equation of the line whose y-intercept is and is parallel to the line x 3y + 7 = Prove that the equation of a straight line passing throngh (a cos 3 θ, a sin 3 θ) and perpendicular to the sine x sec θ + y cosec θ = a is x cos θ y sin θ = a cos θ. 0.8 PPLICTION OF COORDINTE GEOMETRY Let us take some examples to show how coordinate geometry can be gainfully used to prove some geometrical results. Example 0.36 Prove that the diagonals of a rectangle are equal. Solution : Let us take origin as one vertex of the rectangle without any loss of generality and take one side along x axis of length a and another side of length b along y axis, as shown in Fig. 0.. Then the coordinates of the points, and C are (a, 0), (a, b) and (0, b) respectively Length of diagonal C = (a 0) + (0 b) = a + b Length of diagonal O = (a 0) + (0 b) = a + b i.e., C = O the diagonals of the rectangle O C are equal. Thus, in general, we can say that the length of the diagonals of a rectangle are equal. MTHEMTICS 369

32 Y Straight lines C ( 0,b) (a,b) O ()0,0 (a,)0 Fig. 0. X Example 0.37 Prove that angle in a semi circle is a right angle. Solution : Let O (0,0) be the centre of the circle and ( a,0) and (a,0) be the end- points of one of its diameter. Let the circle intersects the y-axis at (0, a) [see Fig. 0.3] Slope of Q = Slope of Q = a 0 0 a = = m 0 a a 0 = = m s m m = Q is perpendicular to Q Q is a right angle. gain, let P (a cos θ, a sin θ) be a general point on the circle. a sin sin Slope of P = = = m cos + a + cos (say) Y Q ( 0,a) ( cos, sin ) P a θ a θ (a, 0) O ()0,0 Fig. 0.3 (a, 0) 370 MTHEMTICS

33 Straight lines 0 a sin sin Slope of Q = = = m a a cos cos (say) and m. m = F HG I KJ F H G sin sin + cos cos P is perpendicular to P. P is a right angle. I K J = sin sin cos = cos = Example 0.38 Medians drawn from two vertices to equal sides of an isosceles triangle are equal. a Solution : Let O be an equilateral triangle with vertices (0,0), (a,0) and, a Let C and D be the mid-points of and O respectively. a y, a Fig. 0.4 D C O (0,0) (a, 0) x C has coordinates D has coordinates Length of OC = F 3a ai, HG 4 K J and F a ai, HG 4 K J F 3 I 9 HG 4 K J + ( ) = + = a a a a a and Length of D = (a a 4 ) + (0 a ) = a 4 3 Length of OC = length of D Hence the proof. MTHEMTICS 37

34 Straight lines Example 0.39 The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it. Solution : Let the coordinates of the vertices be O (0, 0), (a, 0) and (b,.c). Let C and D be the mid-points of O and respectively The coordinates of C are b, c F HG F HG The coordinates of D are a + b, c Slope of CD = I K J c c a + b b = 0 Slope of O, i.e, x axis = 0 CD is parallel to O I K J (0,0) O y C (b,c) D ( a,0) Fig. 0.5 x Length of CD = a + b b c c a + = Length of O = a CD = O which proves the above result. Example 0.40 If the diagonals of a quadrilateral are perpendicular and bisect each other, then the quadrilateral is a rhombus. Solution : Let the diagonals be represented along OX and OY as shown in Fig. 0.6 and suppose that the lengths of their diagonals be a and b respectively. O = OC = a and OD = O = b D = O + OD = a + b ( a,0) D Y ( 0,b) O C ( a,0) X = O + O = a + b (0, b) Fig MTHEMTICS

35 Straight lines Similarly C = a b + = CD s = C = CD = D CD is a rhumbus. CHECK YOUR PROGRESS 0.7 Using coordinate geometry, prove the following geometrical results:. If two medians of a triangle are equal, then the triangle is isosceles.. The diogonals of a square are equal and perpendicular to each other. 3. If the diagonals of a parallelogram are equal, then the paralle logram is a rectangle. 4. If D is the mid-point of base C of C, then + C = (D + D ) 5. In a triangle, if a line is drawn parallel to one side, it divides the other two sides proportionally. 6. If two sides of a triangle are unequal, the greater side has a greater angle opposite to it. 7. The sum of any two sides of a triangle is greater than the third side. 8. In a triangle, the medians pass throngh the same point. 9. The diagonal of a paralleogram divides it into two triangles of equal area. 0. In a parallelogram, the pairs of opposite sides are of equal length. LET US SUM UP The equation of a line parallel to y-axis is x = a and parallel to x-axis is y = b. The equation of the line which cuts off intercept c on y-axis and having slope m is y = mx + c The equation of the line passing through (x, y ) and having the slope m is y y = m (x x ) The equation of the line passing through two points (x, y ) and (x, y ) is y y = y x y x (x x ) The equation of the line which cuts off interceptsa andb onx axis andy axis respectively is x a y + = b The equation of the line in normal or perpendicular form isx cos + y sin = p MTHEMTICS 373

36 Straight lines where p is the length of perpendicular from the origin to the line and is the angle which this perpendicular makes with the positive direction of thex-axis. x x y y The equation of a line in the parametric form is = = r cos sin where r is the distance of any point P (x, y) on the line from a given point (x, y ) on the given line and is the angle which the line makes with the positive direction of the x-axis. The general equation of first degree in x and y always represents a straight line provided and are not both zero simultaneously. From general equation x + y + C = 0 we can evaluate the following : (i) Slope of the line = (ii) x-intercept = C (iii) y-intercept = C (iv) Length of perpendicular from the origin to the line = C ( + ) (v) Inclination of the perpendicular from the origin is given by cos = ( + ) ; sin = ( + ) where the upper sign is taken for C > 0 and the lower sign for C < 0; but if C = 0 then either only the upper sign or only the lower sign are taken. m m If be the angle between two lines with slopes m m then tan = + m m (i) Lines are parallel if m = m. (ii) Lines are perpendicular ifm m =. () ngle between two lines of general form is: tan θ = + (i) Lines are parallel if = 0 or (ii) Lines are perpendicular if + = 0. = 0 Distance of a given point (x,y ) from a given linex + y + C =0 is d = x + y + C ( + ) Equation of a line parallel to the line x + y + C = 0 is x + y + k = 0 Equation of a line perpendicular to the line x + y + C = 0 is x y + k = MTHEMTICS

37 Straight lines SUPPORTIVE WE SITES TERMINL EXERCISE. Find the equation of the straight line whosey-intercept is 3 and which is: (a) parallel to the line joining the points (,3) and (4, 5). (b) perpendicular to the line joining the points (0, 5) and (, 3).. Find the equation of the line passing through the point (4, 5) and (a) parallel to the line joining the points (3, 7) and (, 4). (b) perpendicular to the line joining the points (, ) and (4, 6). 3. Show that the points (a, 0), (0,b) and (3a, b) are collinear. lso find the equation of the line containing them. 4. (, 4), (, 3) and C (, ) are the vertices of triangle C. Find (a) the equation of the median through. (b) the equation of the altitude througle. (c) the right bisector of the side C. 5. straight line is drawn through point (, ) making an angle of 6 with the positive direction of x-axis. Find the equation of the line. 6. straight line passes through the point (, 3) and is parallel to the line x + 3y + 7 = 0. Find its equation. 7. Find the equation of the line having a and b as x-intercept and y-intercepts respectively. 8. Find the angle between the lines y = ( 3 ) x + 5 and y = ( + 3 ) x d. 9. Find the angle between the lines x + 3y = 4 and 3x y = 7 0. Find the length of the per pendicular drawn from the point (3, 4) on the straight line (x + 6) = 5 (y ).. Find the length of the perpendicula from (0, ) on 3x + 4 y + 5 = 0.. Find the distance between the lines x + 3y = 4 and 4x + 6y = 0 MTHEMTICS 375

38 Straight lines 3. Find the length of the perpendicular drawn from the point ( 3, 4) on the line 4x 3y = Show that the product of the perpendiculars drawn from the points on the straight line x y cos + sin θ = is b a b. 5. Prove that the equation of the straight line which passes through the point (a cos 3 θ, b sin 3 θ) and is perpendicular to x secθ + y cosec θ = a is x cosθ - y cosec θ = a cos θ 6. Prove the following geometrical results : (a) (b) The altitudes of a triangle are concurrent. The medians of an equaliateral triangle are equal. (c) The area of an equalateral triangle of side a is 3 4 a (d) If G is the centroid of C, and O is any point, then O + O + OC = G + G + CG + 3OG. (e) The perpendicular drawn from the centre of a circle to a chord bisects the chord. 376 MTHEMTICS

39 Straight lines NSWERS CHECK YOUR PROGRESS 0.. Straight line. (a) (b) (c) (a) y = 4 (b) x = 3 6. x = 5 7. y + 7 = 0 CHECK YOUR PROGRESS 0.. (a) y = x (b) Slope = 4, 3 y intercept =. 3y = 3x 3. Slope =, y-intercept = 4. 3x + 7y = 7 5. y = x + ; x + y 3 = x y = 0 7. (a) x + y = (b) Equation of the diagonal C = x y 4 = 0 Equation of the diagonal D = x y + 66 = 0 8. x = 0, x 3y + 6 = 9 and 5x 3y = 0 9. x + 3y = x + y = 6. 3x + 4y =. x + y = 4. x y = and the co-ordinates of the point are +,+ CHECK YOUR PROGRESS 0.3. and are not both simltaneously zero. y = 3 x (a)x intercept = c m (c) x-intercept = 4; y-intercept = 6 ; y-intercept = c (b) x-intercept = 8 3 ; y-intercept = 8 3 x x 4. 3 units 5. + = p sec p cos ec MTHEMTICS 377

40 6. (a) x + y = 0 (b) The first line is nearer from the origin. CHECK YOUR PROGRESS 0.4. (a) m = F. θ = tan H G I K J (a) (i) (b) Parallel (b) m = 4 x + y = 0 5 = (ii) + = 0 CHECK YOUR PROGRESS 0.5. d = (c) Perpendicular. F I HG K J F 3. 0, 3 I HG 3 K J F 3 I HG 4 ( 4 ± 5), 0 K J b. 0, ( ) a a ± a + b 5. Straight lines CHECK YOUR PROGRESS x + y + = 0. y = x + 3. x y = 3 4. y = x 5. 3x ay = 6 (a ) 6. 4x 3y + 3 = 0 7. x 3y + 6 = 0 TERMINL EXERCISE. (a) 4x + 3y + 9 = 0 (b) x 8y 4 = 0. (a) 3x 5y 37 = 0 (b) 5x 8y 60 = 0 4. (a) 3x y 9 = 0 (b) 3x y + = 0 (c) 3x y 4 = 0 5. x 3y = 3 6. x + 3y +3 = 0 7. bx + ay = ab MTHEMTICS