Tiêu đề XI - maths - chapter 10 - STRAIGHTLINES (52-91).pdf ✅

THE STRAIGHT LINE JEE-MAIN-JR-MATHS VOL-III 52 NARAYANAGROUP  If  is an angle between two nonvertical lines having slopes 1 2 m m, then 1 2 1 2 1 2 tan , 1 1 m m m m m m        i) If  is acute then 1 2 1 2 tan 1 m m m m     ii) If  is one angle between two lines then the other angle is   . Usually the acute angle between two lines is taken as the angle between the lines Intercept(s) of a line:  If a line cuts x-axis at A(a, 0) and y-axis at B(0,b) then a and b are called x-intercept and y-intercept of that line respectively i) Intercept of a line may be positive or negative or zero ii) x-intercept of a horizontal line is not defined iii) y-intercept of a vertical line is not defined iv) Intercepts of a line passing through origin are zero. Equation of a straight line in various forms:  i) Line parallel to x-axis: Equation of horizontal line passing through (a,b) is y = a ii) Line parallel to y-axis: Equation of vertical line passing through (a, b) is x = b iii) Slope - point form :The equation of the line with slope m and passing through the point  x y 1 1 ,  is y– y1 = m (x – x1 ) W.E-2: If (3,-1),(2,4),(-5,7) are the mid points of the sides BC ,CA , AB of triangle ABC. Then the equation of the side CA is Sol :Here m = – 1 and given point (x1 , y1 ) is (2, 4). By point slope form equation of the line is y – 4 = – 1 (x - 2) iv) Two - point form :The equation of a line passing through two points   1 1 2 2 ( , ) , A x y and B x y is       1 2 1 1 2 1 y y x x x x y y      (or) 1 1 2 2 1 1 0 1 x y x y x y  Inclination of a line:  If a line makes an angle        0 with x-axis measured in positive direction then  is called inclination of the line. i) Inclination of horizontal line is zero ii) Inclination of vertical line is  / 2 Slope of a line:  If the inclination of a non vertical line is  then tan is called slope of the line and is usually denoted by m, thus m tan   X Y O  i) Slope of horizontal line (x-axis) is zero   0   0 ii) Slope of vertical line (y-axis) is not defined   0   90 iii) 0     0 0 m ; 0 0 0 90 0      m 0    90 m is not defined 0 0 90 180 0      m  Slope of the line joining two points 1 1 A x y ( , ) , B x y  2 2 ,  is   2 1 1 2 2 1 y y m x x x x     i) If x1 =x2 then the line AB is vertical and hence its slope is not defined ii) If y1 =y2 then the line AB is horizontal and hence its slope is 0  Two nonvertical lines are parallel if their slopes are equal.  Two non vertical lines are perpendicular if product of their slopes is –1 W.E-1:The medians AD and BE of the triangle with vertices A(0,b), B(0,0) and C(a,0) are mutually perpendicular if Sol: AD BE  2 1 b b a a                 ; 2 2   2b a THE STRAIGHT LINE SYNOPSIS
JEE-MAIN-JR-MATHS VOL-III THE STRAIGHT LINE NARAYANAGROUP 53 x + y = xo + yo d) Equation of the line making equal intercepts in magnitude but opposite in sign and passing through (xo ,yo ) is x – y = xo – yo e) The equation of the line passing through the point (x1 , y1 ) and whose intercepts are in the ratio m : n is nx+my=nx1 +my1 (or) mx+ny=mx1 +ny1 W.E-5: The sum of x,y intercepts made by the lines x+y=a, x+y=ar, x+y=ar2 ...... on coordinate axes when r=1/2, a  0 Sol: required sum =   2 2 2 2 ...... infinite . a ar ar G P    = 2a/1-r = 4a vii) General equation of line : a) A linear equation in x and y always represents a line. b) The equation of a line in general form is a x + b y+ c = 0 , where a , b , c are real numbers such that 2 2 a b   0 having slope =-a/b , x-intercept=-c/a , y- intercept =-c/b . c) The equation of a line parallel to ax by c    0 is of the form ax by k    0 , k R  . d) The equation of a line perpendicular to ax by c    0 is of the form bx ay k    0 , k R  e) Equation of a line passing through  x y 1 1 ,  and (i) parallel to ax by c    0 is a x x b y y      1 1    0 (ii)Perpendicular to ax by c   0 is b x x a y y      1 1    0 viii) Normal form : a) The equation of the straight line upon which the length of the normal drawn from origin is 'p' and this perpendicular makes an angle    , 0 2     with positive x-axis is x y p cos sin     ,  p  0 W.E-3: Equation of the diagonal (through the origin) of the quadrilateral formed by the lines x = 0, y = 0, x + y = 1 and 6x + y = 3 is Sol :Here  x y 1 1 , 0,0     ,  2 2  2 3 , , 5 5 x y        Using two-point form, the equation of the line is 3x -2 y = 0 v) Slope - Intercept form : a) The equation of the line whose slope is m and which cuts an intercept ‘c’ on the y-axis is y = mx+c L 0,c O Y X b) The equation of the line whose slope is m and which cuts an intercept ‘a’ on the x-axis is y = m(x - a) c) The equation of the line passing through t h e origin and having slope m is y=mx W.E-4:Equation to the straight line cutting off an intercept 2 from negative y axis and inclined at 30o to the positive direction of axis of x, is Sol :Equation of line passing through (0,-2) and having slope 1 3 is 3y x 2 3 0    vi) Intercept Form :Suppose a line L makes intercept on x-axis is a and on y- axis is b then its equation is x y 1 a b   a) If the portion of the line intercepted between the axes is divided by the point (x1 , y1 ) in the ratio m : n, then the equation of the line is m n y n my 1 1    x x (or) 1 1 mx x ny m n y    b) Equation of the line whose intercept between the axes is bisected at the point (x1 , y1 ) is 2 y y 1 1   x x c) Equation of the line making equal intercepts on the axes and through the point (xo, yo ) is
THE STRAIGHT LINE JEE-MAIN-JR-MATHS VOL-III 54 NARAYANAGROUP X Y Px1  r cos , y1  rsin  Qx1  r cos, y1  rsin  Ax1 , y1 O B   cos  = AP x  x1 , sin  = AP y  y1 or x – x1 = AP cos  , y – y1 = AP sin  . 1 1 cos sin x x y y r       W.E-7: (1,2),(3,6)are two opposite vertices of a rect- angle and if the other two vertices lie on the line 2y = x + c, then c and other two vertices are Sol: Mid point of given vertices is P x y  1 1 , 2,4     which lies on 2y = x + c then c=6. Now r=BP=AP= 5 , 1 tan 2   Hence B= x r y r 1 1   cos , sin    =(4,5) C= x r y r 1 1   cos , sin    =(0,3) Distances:  i) The perpendicular distance to the line ax by c    0 (a) from origin is 2 2 c a b  (b) from the point   1 1 1 1 2 2 , ax by c x y is a b    ii) The distance of a point  x y 1 1 ,  from the line L ax by c    0 measured along a line making an angle  with x-axis is 1 1 cos sin ax by c a b      iii) The distance between parallel lines ax by c   1 0 and ax by c   2 0 is  P X Y L b) The normal form of a line ax by c    0 is     2 2 2 2 2 2 , 0 a b c x y if c a b a b a b         and 2 2 2 2 2 2 , 0 a b c x y if c a b a b a b        W.E-6: Normal form of the equation x+y+1=0 is Sol: The given equation is x+y+1=0-x-y=1  1 1    1 2 2 2   x y    1 cos sin 4 4 2 x y                      5 5 1 cos sin 4 4 2 x y      ix) Symmetric form and Parametric equations of a straight line : a) The equation of the straight line passing through (x1 ,y1 ) and makes an angle  with the positive direction of x-axis is 1 1 cos sin x x y y      Where            0, ( ) b) The co-ordinates  x y,  of any point P on the line at a distance ‘r’ units away from the point   1 1 A x y, can be taken as      1 1 1 1 x r y r x r y r     cos , sin or cos , sin     c) The equations 1 x x r   cos , 1 y y r   sin are called parametric equations of a line with parameter 'r' of the line passing through the point   1 1 x y, and having inclination  .
JEE-MAIN-JR-MATHS VOL-III THE STRAIGHT LINE NARAYANAGROUP 55 iii) A point   1 1 A x y, and origin lies on the same or opposite side of a line L ax by c     0 according as 11 11 c L or c L . 0 . 0   iv) The point   1 1 x y, lies between the parallel lines ax by c ax by c 1 1 2 2       0, 0 or does not lie between them according as 1 1 1 1 1 2 ax by c ax by c     is negative or positive v) The point   1 1 A x y, lies above or below the line L ax by c     0 according as 11 11 0 0 L L or b b   Proof: The fig. Shows a point P(x1 , y1 ) lying above a given line. If an ordinate is dropped from P to meet the line L at N, then the x coordinate of N will be x1 . Putting x = x1 in the equation ax + by + c = 0 gives ordinate of N = – b (ax c) 1  If P(x1 , y1 ) lies above the line, then we have y1 > – b (ax c) 1  i.e. y1 + b (ax c) 1  > 0 i.e. b (ax by c) 1  1  > 0, i.e. b L(x , y ) 1 1 > 0 N P L  ax  by  c  0 Hence, P(x1 , y1 ) lies above the line ax + by + c = 0, and if b L(x , y ) 1 1 < 0, it would mean that P lies below the line ax + by + c = 0. Ceva's Theorem :  If the lines joining any point ‘O’ to the vertices A,B,C of a triangle meet the opposite sides in D,E, F respectively then . . 1  BD CE AF DC EA FB Proof: Without loss of generality take the point P as the origin O. Let A x y B x y C x y  1 1 2 2 3 3 , , , , ,      be the vertices. Slope of AP is 1 1 1 1 0 0 y y x x    1 2 2 2 c c a b   . iv) The distance between the parallel lines ax+by+c1 =0 and ax + by + c2 = 0 measured along the line having inclination  is 1 2 c c a cos b sin     v) The equation of a line parallel and lying midway between the above two lines is 1 2 0 2 c c ax by     vi) Equiation of the line parallel to ax+by+c=0 and at a distance d from the line is 2 2 ax by c d a b      0 W.E-8:The distance between A(2, 3) on the line of gradient 3/4 and the point of intersection P of this line with 5x + 7y + 40 = 0 is Sol : Since m = 3/4, then cos = 4/5 and sin = 3/5. 5 2 7 3 40 355 4 3 41 5 7 5 5 r                    Position of a point (s) w.r.to line (s):  i) The ratio in which the line L ax by c     0 divides the line segment joining     1 1 2 2 A x y and B x y , , is 11 22 L L: where 11 1 1 22 2 2 L ax by c L ax by c       , ii) The points A , B lie on the same side or opposite side of the line L = 0 according as 11 22 L L, have same sign or opposite sign that is 11 22 11 22 L L or L L . 0 . 0   W.E-9:The range of  in the interval (0, )  such that the points (3, 5) and (sin , cos )   lie on the same side of the line x + y – 1 = 0 is Sol :Since 3 5 1 sin cos 1 0                      1 sin 4 2         3 4 4 4 0 2     