 # Basics of Straight Lines Consider a situation when you ask someone about some direction. And they told you to go straight without taking a turn. We are all quite familiar with the word ‘straight’. It means without a bend or not crooked. Today, we will discuss the basics of straight lines.

### Suggested Videos        ## Straight lines

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.

## Basics of Straight Lines

A straight line is the simplest figure in geometry but it forms the most important concept of it.  Where do you find straight lines? A straight road; edge of the ruler, building, pen, pencil; hands of clocks etc. are few examples of it.

Let us discuss other features of a line and the basics of straight lines. In the basics of straight lines, we will learn about slope, the angle of inclination, collinearity, conditions for being parallel or perpendicular lines.  ### 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. How can we find a slope of a line? In the coordinate geometry, if any line l makes an angle θ with the positive direction of 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, θ ≠ 90°. Can we find the slope of a line when coordinates of points on a line are given? ### Slope of a Line When Coordinates of Any Points on the Lines Are Given

Imagine a line l with a slope θ. Two points A (x1, y1) and B (x2, y2) lies on it. The angle of inclination can be acute or obtuse.

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).

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).

### Conditions for Parallelism of Lines in Terms of Slopes

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 ### 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 lhave 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.

## 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 α& m= tan α2

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

## Collinearity of Three Points

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

Problem: What is the slope of the horizontal line and vertical line?
Solution: The slope of horizontal line is zero (m = 0, θ = 0°). The slope of a vertical line is undefined (θ= 90°).

Problem: Find the slope of the line passing through the points (4, 3) and (1, 5).
Solution: slope of the line, m = (y2 − y1)/(x2 − x1) = (5 − 3) / (1 − 4) = −2/3.

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

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