2. KENDRIYA VIDYALAYA IIM LUCKNOW

3. Straight Lines

4. Consider two lines L 1 and L 2 in a coordinate plane with inclinations a 1 and a 2. If α 1 = α 2 ⇒ l 1 l 2 If α 1 ≠ α 2 ⇒ l 1 and l 2 are intersecting lines The intersecting lines L 1 and L 2 form two pairs of vertically opposite equal angles. α 1, α 2 ≠ 90 o

5. Angle between two lines

6. Consider triangle ABC. In ∆ABC: ∠ ABX = ∠ BCA + ∠ BAC (Exterior angle = Sum of interior opposite angles) ⇒ α 2 = α 1 + θ

7. Or θ = α 2 - α 1. Thus, tan θ = tan (α 2 - α 1 ) Or Tan θ = (Tan α 2 - Tan α 1 )/1 + Tan α 1 x Tan α 2 …..(1). Tan α 2 = m 2 (Slope of line l 2 ). Tan α 1 = m 1 (Slope of line l 1 ). Thus: Tan θ = (m 2 - m 1 )/1 + m 1 m 2 ….(2). Now θ + Φ = 180 o (Supplementary angles). ⇒ Φ = 180 o - θ. Tan Φ = tan (180 o - θ). = - tan θ …..(3). ⇒ Tan Φ = - (m 2 - m 1 )/1 + m 1 m 2 ….(4).

8. tan θ = [(m 2 - m 1 )/1 + m 1 m 2 ] ….(2). tan Φ = - [(m 2 - m 1 )/1 + m 1 m 2 ] ….(4). Case I: [(m 2 - m 1 )/1 + m 1 m 2 ] is positive. ⇒ tan θ is positive. ⇒ θ is an acute angle. SOME CASES------ Case I: [(m 2 - m 1 )/1 + m 1 m 2 ] is positive. ⇒ tan Φ is negative. ⇒ Φ is an obtuse angle. Case II: [(m 2 - m 1 )/1 + m 1 m 2 ] is negative. ⇒ tan θ is negative. ⇒ θ is an obtuse angle.

9. Case III: [(m 2 - m 1 )/1 + m 1 m 2 ] is negative. ⇒ tan Φ is positive. ⇒ Φ is an acute angle. tan θ = (m 2 - m 1 )/1 + m 1 m 2. Φ = 180 o - θ.

10. If A, B and C are collinear: Area of ∆ABC = ½ ∣ x 1 (y 2 - y 3 ) + x 2 (y 3 - y 1 ) + x 3 (y 1 - y 2 ) ∣ = 0 Collinearity of points by using the slopes of lines passing through them: Consider two lines AB and BC passing through the given points.

12. Let m 1 and m 2 be the slopes of lines AB and BC, respectively. If A, B and C are collinear: ∠ ABC (θ) = 180 o tan θ = (m 2 - m 1 )/1 + m 1 m 2 ….(1) If θ = 180 o, tan θ = 0 ⇒ (m 2 - m 1 )/1 + m 1 m 2 = 0 Or m 2 - m 1 = 0 Or m 2 = m 1 If three given points are collinear, then the slopes of the lines passing through any two of them are equal.

13. If the slopes of two lines passing through any two of three given points are equal, the given points are collinear. A(2, -1), B(6, 4), C(10, 9) Slope of AB = (y 2 - y 1 )/(x 2 - x 1 ) = {4 - (-1)}/(6 - 2)= 5/4 Slope of BC = (9-5)/(10-6) = 5/4 Slope of AB = Slope of BC = 5/4 Slope of AC = [9 - (-1)]/(10 - 2)= 10/8 = 5/4 Slope of AB = Slope of BC = Slope of AC = 5/4.

14. Parallel and Perpendicular Lines

15. Parallel Lines 14 Two non-vertical lines are parallel if and only if their slopes are equal. If l 1 ║l 2, then m 1 = m 2. If m 1 = m 2, then l 1 ║l 2. l1l1 l2l2 m1m1 m2m2

16. Write the equation of the line that passes through (3,6) and is parallel to y = 2/3x+2. m = 2/3 and the point is (3,6) y = mx+c 6 = 2/3(3)+c 6 = 2+c 4 = c y = 2/3x+4 15

17. Write the equation of the line that passes through (4,-5) and is parallel to y = -2x-4. m = -2 and the point is (4,-5) y = mx+c -5 = -2(4)+c -5 = -8+c 3 = c y = -2x+3 16

18. Write the equation of the line that passes through (-6,4) and is parallel to y=1/3x-1. m=1/3 and the point is (-6,4) y=mx+c 4=(1/3)+c 12-1=c C=11 y =1/3x+6 17

19. Perpendicular Lines 18 Two non-vertical lines are perpendicular if and only if the product of their slopes is - 1. If l 1 ┴l 2, then m 1 ● m 2 = -1. If m 1 ● m 2 = -1, then l 1 ┴l 2. l1l1 l2l2 m2m2 m1m1 Slopes are negative reciprocals

20. Write the equation of the line that passes through (6,-5) and is perpendicular to y = 2x+3. m = -1/2 and the point is (6,-5) y = mx+c -5 = -1/2(6)+c -5 = -3+c -2 = c y = -1/2x-2 19

21. Write the equation of the line that passes through (6,-7) and is perpendicular to y = 2/3x+1. m = -3/2 and the point is (6,-7) y = mx+c -7 = -3/2(6)+c -7 = -9+c 2 = c y = -3/2x+2 20

22. Write the equation of the line that passes through (-4,-3) and is perpendicular to y = x+6. m = -1 and the point is (-4,-3) y=mx+c -3=(-1)4+c -3+4=c -1=c y = -x-7 21

23. Various Forms of the equation of a line 1.Point Slope form 2. Two – Point Form 3. Slope – Intercept form 4. Intercept form 5. Normal form

24. Point Slope Form XX’ Y’ O Y P (x 1, y 1 ) Q (x, y)

25. Two Point Form XX’ Y’ O Y P (x 1,y 1 ) Q (x 2,y 2 )

26. Slope – Intercept Form Consider a line making an angle  with the x- axis and an intercept c with the y-axis Consider a point P (x, y) on it Slope = m = tan  = y=mx+c XX’ Y’ O Y  c Q x y- c L  M

27. Intercept Form XX’ Y’ O Y y P (x, y) x a b B A

28. Normal Form Consider a line meeting the axes at A and B, at a distance p = OQ from the origin making an angle  with the x-axis. Consider a point P (x, y) on this line. Draw PL  OX, LM  OQ and PN  LM.  PLN =  p = OQ = OM + MQ = OM + NP = xcos  +ysin  xcos  +ysin  =p B  p x A