Basics of Straight Lines | Mathematics (Maths) for JEE Main & Advanced PDF Download

Introduction

A line is considered as a geometrical shape with no breadth. It extends in both directions with no endpoints. It is a set of points and only has length. Lines can be parallel, perpendicular, intersecting or concurrent. A line in a coordinate plane forms two angles with the x-axis, which are supplementary.  The angle (say) θ made by the line l with the positive direction of the x-axis and measured anti-clockwise is called the inclination of the line. Thus 0° ≤ θ ≤ 180°.

Slope of a Line

The basics of straight lines start with a slope. 

• A slope is an inclined position. It forms a certain angle with the base. 
• In coordinate geometry, if any line l makes an angle θ with the positive direction of the x-axis, it is called the Inclination of the line.
• The angle is measured in an anti-clockwise way. A slope of a line is the tangent of the inclination, θ i.e., tan θ. It is denoted as m.
  Thus, the slope of a line: m = tan θ, where θ ≠ 90°.

1. When Coordinates of Any Points on the Lines Are Given

• Two points A (x₁, y₁) and B (x₂, y₂) lies on it. 
• The angle of inclination can be acute or obtuse.

(i) Case 1: θ is acute

Here, ∠CAB = θ. The slope of the line, m = tan θ.
In ΔCAB, tan θ = CB/CA = (y₂ − y₁)/(x₂ − x₁).
Thus, m = tan θ = (y₂ − y₁)/(x₂ − x₁).

(ii) Case 2: θ is obtuse

Here, ∠CAB =180° − θ.Slope of the line, m = tan θ = tan (180°− ∠CAB) = − tan ∠CAB
m = − CB/CA = − (y₂ − y₁)/(x₁ − x₂).
Thus, m = tan θ = (y₂ − y₁)/(x₂ − x₁).

2. Conditions for Parallelism of Lines in Terms of Slope

• Two lines are parallel if the distance between them at any point remains the same. 
• It can also be inferred that the slopes of the two lines must be the same. 
• Let two lines l₁ and l₂ have respective slopes m₁ and m₂ and angle of inclinations α and β. 
  The lines will be parallel if α = β i.e., m₁ = m₂ and tanα = tanβ

3. Conditions for Perpendicularity of Lines in Terms of Slopes

• Two lines are perpendicular if they intersect each other at an angle of 90°. 
• Let two lines l₁ and l₂ have respective slopes m1 and m2 and angle of inclinations α and β. 
• Here, α = β + 90°.

• tan α= tan (β + 90°) = − cot β = − 1/tan β
• or,  m₂ = −1 /m₁ or m₁m₂ = −1
• The lines will be perpendicular if and only if m₂ = −1/m₁ or m₁m₂ = −1.

4.  Angle Between Two Lines

• Above we get to know about parallel and perpendicular lines. 
• How can we find out the angle between two intersecting lines (other than 90°)? Let two lines l₁ and l₂ have respective slopes m1 and m2 and angle of inclinations α₁ & α₂. Or, m1= tan α₁ & m₂ = tan α₂

• From the property of angle:
  θ = α₁ − α₂
  tan θ = tan (α1 − α2) 
  = (tanα₁ − tanα₂)/ (1+ tanα₁ tanα₂) 
  = (m₁−m₂)/ (1+m₁m₂)
• Φ = 180° − θ,
  ⇒ tan Φ = tan (180° − θ) 
  = − tan θ 
  = − (m₁−m₂) / (1+m₁m₂)
  1 + m₁m₂ ≠ 0
  

5. Collinearity of Three Points

Three points are collinear if they all lie on the same line. Three points A(x1,y1), B(x2,y2) and C(x3,y3) are collinear iff slope of AB = slope of BC i.e., (y₂ − y₁)/(x₂ − x₁) = (y₃ − y₂)/(x₃ − x₂).

Solved Examples

Question 1: Find the slope of the line passing through the points (2, 3) and (5, 11)

Sol: Given points are (2, 3) and (5, 11)

So, to find the slope when two points are given, use the following formula

m = (y₂ - y₁)/(x₂ - x₁)

m = (11 - 3)/(5 - 2)

m = 8/3

So, the slope of the line passing through the points (2, 3) and (5, 11) is 8/3.

Question 2:Find the slope of the line passing through the points (4, 3) and (1, 5).

Sol: Slope of the line, 
m = (y₂ − y₁)/(x₂ − x₁) 
= (5 − 3) / (1 − 4) 
= −2/3.

Question 3: Find the value of x, if the points (1, −1), (x, 1) and (6, 7) are collinear.

Sol: Three points are collinear if (y₂ − y₁)/(x₂ − x₁) = (y₃ − y₂)/(x₃ − x₂) . 
Putting the values, we have, 
(1 − (−1))/(x − 1) 
= (7 − 1)/(6 − x) or, 2(6 − x)  
= 6(x − 1). 
Or, 
x = 9/4.

Question 4: Find the angle between the following two lines: Line 1: 4x -3y = 8, Line 2: 2x + 5y = 4

Sol: Put  4x -3y = 8 into slope-intercept form so you can clearly identify the slope. 
4x -3y = 8
3y = 4x - 8
y = 4x / 3 - 8/3
y = (4/3)x - 8/3
Put 2x + 5y = 4 into slope-intercept form so you can clearly identify the slope.
2x + 5y = 4
5y = -2x + 4y = -2x/5 + 4/5
y = (-2/5)x + 4/5
The slopes are 4/3 and -2/5 or 1.33 and -0.4. 
It does not matter which one is m₁ or m₂. 
You will get the same answer.
Let m₁  = 1.33 and m₂ = -0.4
 tan θ = ± (m1 – m₂ ) / (1+ m₁m₂) 
tan θ = ± (1.33 - (- 0.4)) / (1- (1.33)(-0.4)) 
tan θ = ± (1.73) / (1- 0.532) 
tan θ = ± (1.73 ) / (0.468) 
tan θ= 3.696θ 
tan⁻¹ (3.69)

Question 5: Find the acute angle between y = 3x+1 and y = -4x+3

Sol: m₁= 3 and m₂ = -4 
tan θ = ±  (m1 – m2 ) / (1+ m₁m₂) 
tan θ = ± (3-(-4) ) / (1+ 3(-4)) 
tan θ = ±  (7 ) / (1+(-12)) 
tan θ = ± (7 ) / (-11) 
tan θ = ± (7/11) 
tan θ = 0.636θ 
tan⁻¹ (0.636)