KENDRIYA VIDYALAYA

IIM LUCKNOW

Straight Lines

Consider two lines L1 and L2 in a coordinate plane with inclinations a1 and a2.If α1 = α2 ⇒ l1   ⃦ l2If α1 ≠ α2 ⇒ l1 and l2 are intersecting linesThe intersecting lines L1 and L2 form two pairs of vertically opposite equal angles.α1, α2 ≠ 90o

Angles between two points

Angle between two lines

Theorem

If is acute angle between two non vertical and non-perpendicular lines L1 and L2 with

slopes m1 and m2 i.e.,

tan

Consider triangle ABC.In ∆ABC: ∠ABX = ∠BCA + ∠BAC (Exterior angle = Sum of interior opposite angles)⇒ α2 = α1 + θ

Or θ = α2 - α1.

Thus, tan θ = tan (α2 - α1) Or Tan θ = (Tan α2 - Tan α1)/1 + Tan α1 x Tan α2 …..(1).Tan α2 = m2 (Slope of line l2).Tan α1 = m1 (Slope of line l1).

Thus:Tan θ = (m2 - m1)/1 + m1m2 ….(2).Now θ + Φ = 180o (Supplementary angles).⇒ Φ = 180o - θ.Tan Φ = tan (180o - θ).= - tan θ …..(3).⇒ Tan Φ = - (m2 - m1)/1 + m1m2 ….(4).

tan θ = [(m2 - m1)/1 + m1m2] ….(2).tan Φ = - [(m2 - m1)/1 + m1m2] ….(4).Case I: [(m2 - m1)/1 + m1m2] is positive.⇒ tan θ is positive .⇒ θ is an acute angle.

SOME CASES------

Case I: [(m2 - m1)/1 + m1m2] is positive.⇒ tan Φ is negative.⇒ Φ is an obtuse angle.Case II: [(m2 - m1)/1 + m1m2] is negative.⇒ tan θ is negative .⇒ θ is an obtuse angle.

Case III: [(m2 - m1)/1 + m1m2] is negative.⇒ tan Φ is positive.⇒ Φ is an acute angle.

tan θ = ⎢(m2 - m1)/1 + m1m2 ⎢.Φ = 180o - θ.

If A, B and C are collinear: Area of ∆ABC = ½ ∣x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)∣ = 0Collinearity of points by using the slopes of lines passing through them:Consider two lines AB and BC passing through the given points.

COLLINEARITY

Let m1 and m2 be the slopes of lines AB and BC, respectively.If A, B and C are collinear: ∠ABC (θ) = 180o

tan θ = (m2 - m1)/1 + m1m2 ….(1)If θ = 180o, tan θ = 0⇒ (m2 - m1)/1 + m1m2 = 0Or m2 - m1 = 0Or m2 = m1

If three given points are collinear, then the slopes of the lines passing through any two of them are equal.

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 = (y2 - y1)/(x2 - x1) = {4 - (-1)}/(6 - 2)= 5/4Slope of BC = (9-5)/(10-6) = 5/4Slope of AB = Slope of BC = 5/4Slope of AC = [9 - (-1)]/(10 - 2)= 10/8 = 5/4Slope of AB = Slope of BC = Slope of AC = 5/4.

Parallel and Perpendicular

Lines

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Parallel Lines

• Two non-vertical lines are parallel if and only if their slopes are equal.• If l1║l2, then m1= m2.

• If m1= m2, then l1║l2.

l1 l2

m1 m2

15

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

16

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

17

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

Perpendicular Lines

18

• Two non-vertical lines are perpendicular if and only if the product of their slopes is -1.• If l1┴l2, then m1 ● m2= -1.

• If m1● m2 = -1, then l1┴l2.

l1 l2

m2m1

Slopes are negative

reciprocals

19

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

20

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

21

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

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

Point Slope Form

Consider a line passing through P (x1, y1) and having a slope m.

Consider any point Q (x, y) on it.

slope m =

y-y1=m(x-x1)

XX’

Y’

O

Y

P (x1, y1)

Q (x, y)

Two Point Form

Consider a line passing through P (x1,y1) and Q (x2,y2).

slope m =

Using point slope form,

y-y1= (x-x1) XX’

Y’

O

Y

P (x1,y1)

Q (x2,y2)

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+cXX’

Y’

O

Y

c

Qx

y-c

L

M

PMQM

y cx

Intercept Form

Consider a line making intercepts a and b on the axes.

Consider a point P (x, y) on it.

Area of OPB + Area of OPA = Area of OAB

XX’

Y’

O

Y

y

P (x, y)x

a

b

B

A

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, y)

p

x

A