Straight Lines Class 11th Notes - Free NCERT Class 11 Maths Chapter 10 notes - Download PDF

If you are told to go from one end of a cricket ground to the other end, what would you do? You would generally take the shortest path, right? Well, have you noticed that the shortest path between two points is always a straight line? Straight lines are one of the most fundamental parts of geometry, which play a significant role in navigation, architecture, road construction, etc. From NCERT Class 11 Maths, the chapter Straight Lines contains the concepts of Slope of a line, Angle between two lines, Different forms of the equation of a line, Distance of a point from a line, etc. Understanding these concepts will help the students grasp more advanced trigonometry topics easily and will also enhance their problem-solving ability in real-world applications.

This article on NCERT notes Class 11 Maths Chapter 9 Straight Lines offers well-structured NCERT notes to help the students grasp the concepts of Straight Lines easily. Students who want to revise the key topics of Straight Lines quickly will find this article very useful. It will also boost the exam preparation of the students by many folds. These notes of NCERT Class 11 Maths Chapter 9 Straight Lines are made by the Subject Matter Experts according to the latest CBSE syllabus, ensuring that students can grasp the basic concepts effectively. NCERT solutions for class 11 maths and NCERT solutions for other subjects and classes can be downloaded from the NCERT Solutions.

NCERT Class 11 Maths Chapter 9 Notes

Important formulas:

1. The distance between two points $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ and $\mathrm{Q}({\mathrm{x}}_2,{\mathrm{y}}_2)$ is $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
2. The coordinate of point which cuts a line segment joining by $P(x_1,y_1)$ and $Q(x_2,y_2)$ internally in ratio $m:n$ is $(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$
3. If the ratio of $\mathrm{m}:\mathrm{n}=1:1$ then the coordinate of the point is $(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})$
4. Area of the triangle whose vertices are $P(x_1,y_1),Q(x_2,y_2)$, and $R(x_3,y_3)$ is given by:
$$A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$$

Slope Of A Line

The steepness of a line is known as the slope of the line.

A line that is nonparallel to the x-axis cuts the x-axis to two angles. The angles are supplementary of each other. A line that makes an angle with the positive direction x-axis and the angle is measured anti-clockwise, then θ is the inclination of the line.

Definition: If a line makes an angle θ with the positive direction x-axis, then tanθ is slope of the line.

The slope of a line is also denoted by the letter m,

Slope of a line when coordinates of any two points on the line are given

When two points are given as (x₁, y₁) and (x₂, y₂) the slope is $m=\frac{y_2-y_1}{x_2-x_1}$

Conditions for Parallelism and Perpendicularity of Lines in Terms of Their Slopes

Parallel Lines

Let two lines l₁ and l₂ be parallel to each other. The slopes of lines l₁ and l₂ are m₁ and m_2. The inclinations of the lines l₁ and l₂ are α and β, respectively. The diagram is shown below

Since the lines are parallel to each other so the angle of inclination is also equal.

Hence, $\alpha =\beta$

Taking $\tan$ both sides

$\tan \alpha =\tan \beta$ and m₁ = m₂

The slope of line l₁ = The slope of line l₂

If the lines are parallel, then the slopes of the lines are also equal.

Perpendicular Lines

Let two lines l₁ and l₂ be perpendicular to each other. The slopes of lines l₁ and l₂ are m₁ and m₂. The inclinations of the lines l₁ and l₂ are α and β, respectively. The diagram is shown below

So, $\beta =\alpha +90°$

Taking $\tan$ both sides

$\begin{array}{rl}& \tan \beta =\tan (\alpha +{90}^{\circ })\\ & \Rightarrow \tan \beta =-\cot \alpha \\ & \Rightarrow \tan \beta =-\frac{1}{\tan \alpha }\\ & \Rightarrow \tan \alpha \cdot \tan \beta =-1\\ & \Rightarrow m_1{m}_2=-1\end{array}$

Two lines l1 and l2 are said to be perpendicular if m₁m₂ = -1

The Angle Between Two Lines

The inclination of two lines be θ₁ and θ₂ and θ₁, θ₂ ≠ 90°

The slopes of the lines are m₁ = tanθ₁ and m₂ = tanθ₂

Assume that θ is the angle between the lines.
So,
$\begin{array}{rl}\tan \theta & =\tan ({\theta }_1-{\theta }_2)\\ & =\frac{\tan {\theta }_1-\tan {\theta }_2}{1+\tan {\theta }_1\tan {\theta }_2}\\ & =\frac{m_1-m_2}{1+m_1{m}_2}\end{array}$

Example 3.4
Find the acute angle between the lines $2x-y+3=0$ and $x+y+2=0$
Solution
Let $m_1$ and $m_2$ be the slopes of $2x-y+3=0$ and $x+y+2=0$
Now $m_1=2,m_2=-1$
Let $\theta$ be the angle between the given lines
$\begin{array}{rl}\tan \theta & =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|\\ & =\left|\frac{2+1}{1+2(-1)}\right|=3\\ \theta & ={\tan }^{-1}(3)\end{array}$

Collinearity Of Three Points

Let three points be A(1,4), B(4,6), and C(10,10) lies on the same line.

The slope of the line that passes through the points $A(1,4)$ and $B(4,6)$ is $\frac{6-4}{4-1}=\frac{2}{3}$

The slope of the line that passes through points $B(4,6)$ and $C(10,10)$ is $\frac{10-6}{10-4}=\frac{2}{3}$

So, m_AB = m_BC = $\frac{2}{3}$

If the points lie on the same line, then the slope of the line joining any two points is always the same.

Various Forms of the Equation of a Line

Horizontal and vertical lines

If a line is parallel to the x-axis and the distance of the line from the x-axis is b units. Then the equation of the line is

Either $y=b$ or $y=-b$.

If a line is parallel to the y-axis and the distance of the line from the y-axis is a unit. Then the equation of the line is

Either $x=a$ or $x=-a$

Point-Slope Form

Let a fixed point be (x₁,y₁) and an arbitrary point be (x,y).

Let m be the slope of the line.

Then $\frac{y-y_1}{x-x_1}=m$

$y-y_1=m(x-x_1)$

Two-point form

Let two points be (x₁,y₁) and (x₂,y₂) then

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Slope-Intercept Form

Let the point on the y-axis be (0,b), and m is the slope of the line

Then the formula is $y=mx+b$

Intercept - Form

Let the point on the coordinate axes be (a, 0) and (0, b)

Then the equation

$\frac{x}{a}+\frac{y}{b}=1$

Normal Form

$x\cos \alpha +y\sin \alpha =p$

General Equation Of A Line

The general equation goes by $Ax+By+C=0$

Different forms of $Ax+By+C=0$

Slope-intercept form

$\begin{array}{rl}& By=-Ax-C\\ & y=-\frac{A}{B}x-\frac{C}{B}\\ & m=-\frac{A}{B}\text{ and }c=-\frac{C}{B}\end{array}$

Intercept form

$\begin{array}{rl}& Ax+By+C=0\\ & \Rightarrow Ax+By=-C\\ & \Rightarrow \frac{x}{-\frac{C}{A}}+\frac{y}{C}=1\\ & \Rightarrow \frac{x}{a}+\frac{y}{b}=1\\ & \Rightarrow a=-\frac{C}{A}\text{ and }b=-\frac{C}{B}\end{array}$

Distance of a Point from a Line

Let P(x₁, y₁) be a point that does not lie on the line $Ax+By+C=0$.

Then the distance of the point P(x₁, y₁) from the line is

$d=\left|\frac{A{x}_1+B{y}_1+C}{\sqrt{A^2+B^2}}\right|$

Distance Between Two Parallel Lines

Suppose the equations of two parallel lines be $Ax+By+C_1=0$ and $Ax+By+C_2=0$

Then the distance between two parallel lines is $d=\left|\frac{C_1-C_2}{\sqrt{A^2+B^2}}\right|$

Importance of NCERT Class 11 Maths Chapter 9 Notes

NCERT Class 11 Maths Chapter 9 Notes play a vital role in helping students grasp the core concepts of the chapter easily and effectively, so that they can remember these concepts for a long time. Some important points of these notes are:

• Effective Revision: These notes provide a detailed overview of all the important theorems and formulas, so that students can revise the chapter quickly and effectively.
• Clear Concepts: With these well-prepared notes, students can understand the basic concepts effectively. Also, these notes will help the students remember the key concepts by breaking down complex topics into simpler and easier-to-understand points.
• Time Saving: Students can look to save time by going through these notes instead of reading the whole lengthy chapter.
• Exam Ready Preparation: These notes also highlight the relevant contents for various exams, so that students can get the last-minute minute very useful guidance for exams.

NCERT Class 11 Notes Chapter Wise

Subject-Wise NCERT Exemplar Solutions

After finishing the textbook exercises, students can use the following links to check the NCERT exemplar solutions for a better understanding of the concepts.

• NCERT Exemplar Class 11 Solutions
• NCERT Exemplar Class 11 Maths
• NCERT Exemplar Class 11 Physics
• NCERT Exemplar Class 11 Chemistry
• NCERT Exemplar Class 11 Biology

Subject-Wise NCERT Solutions

Students can also check these well-structured, subject-wise solutions.

• NCERT Solutions for Class 11 Mathematics
• NCERT Solutions for Class 11 Chemistry
• NCERT Solutions for Class 11 Physics

• NCERT Solutions for Class 11 Biology

NCERT Books and Syllabus

Students should always analyze the latest CBSE syllabus before making a study routine. The following links will help them check the syllabus. Also, here is access to more reference books.

• NCERT Book for Class 11
• NCERT Syllabus for Class 11

Happy learning !!!