LearnPick Navigation
Close

Study Material On Straight Lines

Published in: AIEEE | IIT JEE Mains | Mathematics
144 views
  • Dipu T

    • Guwahati
    • 8 Years of Experience
    • Qualification: B.Tech/B.E.
    • Teaches: Science, Physics, NTSE, Mathematics, Chemistry, II...
  • Contact this tutor

This booklet consists of basic definitions, different forms of straight lines, formulas and practice questions on the discussed topics.

  • 1
    Compiled by : Dipu Sir STRAIGHT LINES BASIC CONCEPTS 1. 2. 3. 4. The Straight Liné : The every equation of first degree in x and y represents a straight line. Slope of a Line : The slope of a line is the tangent of the angle, measured anticlockwise from the positive direction of x-asix, to the line. If the inclination of the line is 0, then slope of the line, (m) tan 0. Different Forms of Staight Line : O (i) Euation of axes : The equation of x-axis is y = 0 The equatioq of y-axis is x = 0. 0 o x x (ii) Equation of straight lines parallel to axes : The equation of the straight line parallel to x-axis and at a distance of b units from it is y = b The equation of a straight line parallel to y-axis and at a distance of 'a' units it is, x = a. (iii) : The equation of a straight line which makes an angle 0 with the x-axis and which has intercept c on the y-axis is y = mx + c, where m = tan 6. (iv) Intercept from : The equation of a straight line making intercepts a and b on x-axis and y-axis respectively, is 1+1=1 (v) Normal from : The equation of a straight line which has a perpendicular makes an angle a with the positive direction of x-axis is x cos a + y sin a = p (vi) One-point form : The equation of a straight line passing through Yl) having gradient mis y—Yl = (vii) Two-point from : The equation of a straight line passing through the points (XP Yl) and (x? Y2) is Y-YI = x (viii) Distance from : The equation of a straight line passing through Yl) and making and angle 9 with x-axis 1 is =r where r is the distance of any point on the line from the point (XP y,) cos0 sino Hence the coordinates of any point on the line are given by x = x I + r cos e and y = yr + r sin 0. Angle between two lines : Let the equations of two straight lines be y m x + c , and ml = tan Or, rn2 = tan Or Let 9 be the angle between the two straight lines. 01 = 0+02 —O 2 tan 0 = tan (91 — 02) tano — tano 2 or tan 0 = 1+ tano tano 1+mm 2 12 If lines are parallel, then 0 = 0 2 = 0 ml — rn2 = 0 ml = m2 I + mm 12 2 o 1 x
  • 2
    If lines are parallel, then 6 = 2 2 I + m mi mime = —1. 1+mm 12 5. Condition for the concurrency of three straight lines : 1st Mathod : Find the point of intersection of any two lines and show that it satisfies the third also. 2nd Method : If P = 0, Q = 0, and R = 0 equations of the given lines and if IP + mQ + nR = 0 takes the form ox + oy + oz = 0 where l, m and n are any three constants to be found by inspection then the three given lines are concurrent. 3rd Method : The three straight lines atx + bjy + ct = 0, a2X + b2Y + = O and + b3Y + = O a 1 are concurrent if a 2 b 1 2 b 3 1 3 6. 7. 8 Perpendicular distance of a point from a line : The Peropendicular distance of a point (XP Yl) from the line ax + by +c ax + by + c = 0 is given by Bisector of angles : The equations of the straight lines bisecting the angles between the lines ax + b y +c 1 all + b y + = 0 and + b2Y + = 0 are 2 1 Note : If in (i), cr and cr are of same sign, then (a). The angle between the given lines in which origin lies is acute, if ala2 + blb2 > 0 (b) The positive sign of R.HS. of (i) will give the bisector of Acute angle, if ala 2 + bLb2 < 0 obtuse angle, if ala 2 + bib2 > 0 2 2 2 Image (x, y) of point (XP y,) about the line ax + by + c = 0 is given by 2 (ax + by +c) a b and foot (a, B) f point (x p Yl) on the line ax + by + c = 0 is given by 13 —Y (ax + by +c) (i) ax + by + c a b
  • 3
    9. Equation of line making angle a from a line of slope tan O and passing through (h, Yl) is y axis B(x, y) A(xt, YD x axis y —Yl = tan (0 ± a) (x — Xi) Locus Locus is set of points, which satisfy given condition (s). General method to find the læus : (i) Take the co-ordinates of P as (h, k), (a, ß) or (x I, Yl) (ii) Unisng the equations given / the conditions given, find an equation in h and k by eliminating the various paramenters given. Generalise (h, k) by replacing h by x and k by y. The equation thus obtained is the locus of the required point p. Position of a point (xt, Yl) with respect to a line ax + by + c = 0 The point (xj, Yl) lies on the same side or opposite side of a line ax + by + c = 0 ax I + by I +c > 0 or
  • 4
    ml + and b a b Point of Intersection of Lines Let S ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 represents a pair of straght lines, then differentiate it with respect to x (assuming y constant) = 2ax + 2hy + 2g = 0 Now differentiate it with respect to y•(assuming x constant) — = 2by + 2hx + 2f = O (i) (ii) Solution of equation (i) and (ii) gives the intersection point of lines. Angle between Pair of straight Lines The angle 0 between the lines represented by S = ax2 + 2hxy+ by2 = 0 and is given by tan e = a+b (i) If a + b = 0, then lines are perpendicular. (ii) If 112 = ab, then lines are parallel. Slope of Lines S ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 Consider only two degree homogeneous part of this equation, i.e., ax2 + 2hxy + by2 = 0 2 or b x (on dividing it by x2) x (Here — represent slopes of lines) x Equation of Angle Bisectors Equation of angle bisectors of the angles between the lines represented by the equation x2—y2 _ xy ax2 + 2hxy + by2 = 0 is a—b h Point of Intersection of the Lines The point of intersection of bg—hf af—gh ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is h2-ab' h2-ab Homogenization Let curve is ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 Now line Ix + my = I intersect it at two points A and B, then equation of pair of straight lines passing through origin and intersection points of curve and line AB i.e. OA and 0B is ax2 + by2 + 2hxy + 2gx (Ix + my) + 2fy (Ix + my) + c (Ix + my)2 y axis o x axis
  • 5
    Angle between the lines which do not pass through origin : The angle between the lines represented by the equation ax2 +by2 + 2hxy + 2gx + 2fy + c = 0 is given by -1 e = tan a+b The lines are perpendicular, if a + b = 0 The lines are parallel, if hi = ab Translation of Axes Transformation of axis is done to reduce calculation. It is of two types : (i) Tranilation of Axis : In this, origin is shifted and that changing the direction of axes by using the substitution . (ii) Rotation of Axes : In this, axes are rotated through some angle, say, 0 by using the substitutions x = X cos 0 -- Y sin 0 y = X sin 0 + Y cos 6 Important Points Area of triangle whose sides are arx +bry + cr = 0, r = l, 2, 3 -2 0 0 x x x x where q, C and C3 ae cofactors of q, and respectively in the determinant is 21CCCl 123 2 a 3 b 1 b 2 b 3 a 1 c 1 •c 2 b 1 b 2 b 3 2 c 1 c 2 = I(alb2 — a2bl) (a2b3 a3b2) (agbl — atb3)l A point (x] , h) lies between two parillel lines ax + by + = 0 and ax + by + c2 = 0 if (ax I + by I + q) (ax I + byl + % ) < () Equation of line parallel to lines ax + by + = 0 and ax + by + = 0 and equidistant from the lines is 2 = 0 as m: n = 1 : I ax + by + 2

Discussion

Copyright Infringement: All the contents displayed here are being uploaded by our members. If an user uploaded your copyrighted material to LearnPick without your permission, please submit a Takedown Request for removal.

Need a Tutor or Coaching Class?

Post an enquiry and get instant responses from qualified and experienced tutors.

Post Requirement

Related Notes

Query submitted.

Thank you!

Drop Us a Query:

Drop Us a Query