Straight Lines Class 11 Notes for CBSE and JEE

Straight Lines is Chapter 9 of NCERT Class 11 Mathematics, and Conic Sections is Chapter 10. Straight lines are the base of coordinate geometry: the same slope, distance and family-of-lines ideas return in circles, conic sections and three-dimensional geometry. These Straight Lines Class 11 Notes follow the NCERT chapter, and they add the JEE Main topics that the textbook does not cover in depth.

India's Best Exam Preparation for Class 11th - Download Now

Straight Lines Class 11 Notes

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

Straight Lines Class 11 Notes: Chapter Overview and Syllabus

The NCERT chapter covers:

• Slope of a line and the angle between two lines
• Horizontal and vertical lines
• Point-slope form, two-point form, slope-intercept form, intercept form and normal form
• General equation of a line
• Distance of a point from a line, and the distance between parallel lines
• Collinearity of points

The JEE Main syllabus (coordinate geometry) also lists the intersection of lines, conditions for the concurrence of three lines, the bisectors of the angles between two lines, the centroid, orthocentre and circumcentre of a triangle, and the family of lines through the point of intersection of two lines. Check the latest syllabi of CBSE and NTA for the exact lists.

Tip from the eSaral Maths team: Draw a rough figure for every problem, and mark the slope or the intercepts on it. A small diagram catches most sign mistakes before you calculate.

Basic Formulas Before Straight Lines in Straight Lines Class 11 Notes

For points $A(x_1, y_1)$ and $B(x_2, y_2)$:

$$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Section formula (internal division in the ratio $m : n$) and midpoint:

$$\left(\frac{m x_2 + n x_1}{m + n},\ \frac{m y_2 + n y_1}{m + n}\right), \qquad \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)$$

Centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$:

$$G = \left(\frac{x_1 + x_2 + x_3}{3},\ \frac{y_1 + y_2 + y_3}{3}\right)$$

Area of a triangle:

$$\text{Area} = \frac{1}{2}\,\Big\lvert\, x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\,\Big\rvert$$

Slope of a Line in Straight Lines Class 11 Notes

The slope of a line is $m = \tan\theta$, where $\theta$ is the angle that the line makes with the positive $x$-axis, measured anticlockwise ($0^\circ \le \theta < 180^\circ$, $\theta \neq 90^\circ$).

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

• Parallel lines: $m_1 = m_2$
• Perpendicular lines: $m_1 m_2 = -1$

Angle Between Two Lines in Straight Lines Class 11 Notes

If two lines have slopes $m_1$ and $m_2$, the acute angle $\theta$ between them is given by:

$$\tan\theta = \left\lvert \frac{m_1 - m_2}{1 + m_1 m_2} \right\rvert$$

For the lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

Forms of the Equation of a Line in Straight Lines Class 11 Notes

For the general form $Ax + By + C = 0$: the slope is $-\dfrac{A}{B}$, the $x$-intercept is $-\dfrac{C}{A}$ and the $y$-intercept is $-\dfrac{C}{B}$.

In the normal form, $p > 0$ is the length of the perpendicular from the origin, and $\omega$ is the angle that this perpendicular makes with the positive $x$-axis.

Symmetric (distance) form, for JEE: a line through $(x_1, y_1)$ at angle $\theta$ is given by:

$$\frac{x - x_1}{\cos\theta} = \frac{y - y_1}{\sin\theta} = r$$

where $r$ is the distance of the point $(x, y)$ from $(x_1, y_1)$.

Distance Formulas in Straight Lines Class 11 Notes

Distance of a point $(x_1, y_1)$ from the line $Ax + By + C = 0$:

$$d = \frac{\lvert A x_1 + B y_1 + C \rvert}{\sqrt{A^2 + B^2}}$$

Distance between the parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$:

$$d = \frac{\lvert C_1 - C_2 \rvert}{\sqrt{A^2 + B^2}}$$

To use the second formula, write both equations with the same coefficients of $x$ and $y$ first.

Collinearity and Concurrency in Straight Lines Class 11 Notes

• Three points are collinear if the area of the triangle formed by them is zero, that is, $x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$. You can also check that the slopes of $AB$ and $BC$ are equal.
• Three lines are concurrent if they pass through one point. For the lines $a_ix + b_iy + c_i = 0$ ($i = 1, 2, 3$), the condition is:

$$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$$

Family of Lines and Angle Bisectors for JEE in Straight Lines Class 11 Notes

Family of lines: every line through the point of intersection of $L_1 = 0$ and $L_2 = 0$ is of the form

$$L_1 + \lambda L_2 = 0$$

where $\lambda$ is a parameter. Use it when the intersection point is not easy to find, or when one more condition fixes $\lambda$.

Bisectors of the angles between two lines:

$$\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm\,\frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$$

The two signs give the two bisectors. To find the bisector of the angle that contains the origin, make $c_1$ and $c_2$ positive first, and use the plus sign.

Triangle centres: the centroid is $G = \left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right)$. The orthocentre, the centroid and the circumcentre lie on one line, and the centroid divides the line from the orthocentre to the circumcentre in the ratio $2 : 1$.

Key Formulas in Straight Lines Class 11 Notes

Solved Examples from Straight Lines Class 11 Notes

Example 1 (slope and angle): Find the slope of the line through $(1, 2)$ and $(4, 8)$, and the angle it makes with the $x$-axis.

$m = \dfrac{8 - 2}{4 - 1} = 2$, so $\tan\theta = 2$ and $\theta \approx 63.4^\circ$.

Example 2 (point-slope form): Find the equation of the line through $(2, 3)$ with slope $4$.

$y - 3 = 4(x - 2)$, which gives $4x - y - 5 = 0$.

Example 3 (intercept form): Find the equation of the line that cuts off intercepts $3$ on the $x$-axis and $-2$ on the $y$-axis.

$\dfrac{x}{3} + \dfrac{y}{-2} = 1$, which gives $2x - 3y = 6$.

Example 4 (distance of a point from a line): Find the distance of $(1, 2)$ from the line $3x + 4y - 10 = 0$.

$d = \dfrac{\lvert 3(1) + 4(2) - 10 \rvert}{\sqrt{9 + 16}} = \dfrac{1}{5} = 0.2$.

Example 5 (angle between lines): Find the acute angle between $y = 2x + 1$ and $y = 3x - 2$.

$\tan\theta = \left\lvert \dfrac{2 - 3}{1 + 6} \right\rvert = \dfrac{1}{7}$, so $\theta = \tan^{-1}\dfrac{1}{7} \approx 8.1^\circ$.

Example 6 (collinearity): Show that $A(1, 2)$, $B(2, 4)$ and $C(3, 6)$ are collinear.

$\text{Area} = \dfrac{1}{2}\lvert 1(4 - 6) + 2(6 - 2) + 3(2 - 4) \rvert = \dfrac{1}{2}\lvert -2 + 8 - 6 \rvert = 0$, so the points are collinear.

Example 7 (distance between parallel lines): Find the distance between $3x + 4y + 5 = 0$ and $3x + 4y - 10 = 0$.

$d = \dfrac{\lvert 5 - (-10) \rvert}{\sqrt{9 + 16}} = \dfrac{15}{5} = 3$.

Example 8 (concurrent lines): Show that $x + y - 3 = 0$, $2x - y = 0$ and $x - 2y + 3 = 0$ are concurrent.

Solving the first two lines gives $3x = 3$, so $x = 1$ and $y = 2$. Substituting $(1, 2)$ in the third line gives $1 - 4 + 3 = 0$, so all three lines pass through $(1, 2)$.

Common Mistakes in Straight Lines Class 11 Notes Problems

How to Study Straight Lines Class 11 Notes

1. Read the NCERT chapter first, and solve its exercises. eSaral’s NCERT Solutions for Class 11 Maths Chapter 9 give the answers to Exercises 9.1 to 9.3.
2. Write a one-page sheet of the forms of a line and the distance formulas, and practise converting between the forms.
3. Practise the JEE topics (families of lines, bisectors and concurrency) after the NCERT exercises.
4. Revise other formula sheets too, such as the trigonometry formulas for Class 11, and the Class 11 Maths notes for all chapters.