Home » Maths » Straight Lines Short Notes & Formula for JEE Main and Advanced

# Straight Lines Short Notes & Formula for JEE Main and Advanced

## Distance Formula

For two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ in the Cartesian plane, the distance between them is given by:

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

## Section Formula

For two points $A(x_1, y_1)$ and $B(x_2, y_2)$, the coordinates ( (x, y) ) of a point ( P ) dividing the line segment ( AB ) internally in the ratio ( m:n ) are:

$x = \frac{m x_2 + n x_1}{m + n}$

$y = \frac{m y_2 + n y_1}{m + n}$

For two points $A(x_1, y_1)$ and $B(x_2, y_2)$, the coordinates ( (x, y) ) of a point ( P ) dividing the line segment ( AB ) externally in the ratio ( m:n ) are:

$x = \frac{m x_2 – n x_1}{m – n}$

$y = \frac{m y_2 – n y_1}{m – n}$

## Mid Point Formula

For two points $A(x_1, y_1)$ and $B(x_2, y_2)$, the coordinates ( (x, y) ) of a mid point ( P ) is :

$x = \frac{x_2 + x_1}{2}$

$y = \frac{y_2 + y_1}{2}$

## Area of Triangle

Area of triangle ABC of coordinates $A(x_1, y_1)$ , $B(x_2, y_2)$ and (\ C(x_3, y_3) \)

$A=\frac {1}{2}|x_1 (y_2-y_3 )+x_2 (y_3-y_1 )+x_3 (y_1-y_2 )|$

## Formula For Collinearity

For point A, B and C to be collinear, the value of A should be zero

$|x_1 (y_2-y_3 )+x_2 (y_3-y_1 )+x_3 (y_1-y_2 )|=0$

## Definition of Slope of the Straight line

• If $\theta$ is the inclination of a line l, then $\tan \theta$ is called the slope or gradient of the line l.
• The slope of a line whose inclination is 90° is not defined.
• The slope of a line is denoted by m.
• Thus, $m = \tan \theta$, $\theta \ne 90$

## Slope Formula

$m = \frac{y_2 – y_1}{x_2 – x_1}$

## Angle Between Two Lines

If $m_1$ and $m_2$ are the slopes of two lines, then the angle ( $\theta$ ) between them is:

$\tan \theta =| \frac{m_2 – m_1}{1 + m_1 m_2}|$

If the lines are parallel, then $m_1=m_2$

if the lines are perpendicular $m_1m_2 =-1$

## General Equation of Straight line

An equation of the form 0, $ax+ by+ c=0$ where a, b, c are constants and a, b are not simultaneously zero, always represents a straight line.

## Various form Equation of Straight line

### Point-slope form

For a line with slope ( m ) passing through the point $(x_1, y_1)$

$y – y_1 = m(x – x_1)$

### Two-point form

For a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$

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

### Slope Intercept Form

For a line with slope ( m ) and which cuts an intercept c on the y-axis

$y=mx +c$

### Intercept Form

For a line intercepting the x-axis at ( a ) and the y-axis at ( b )

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

### Normal Form

For a line making an angle $\omega$ with the positive x-axis and having a perpendicular distance ( p ) from the origin:

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

In normal form of equation of a straight line p is always taken as positive and ? is measured from positive direction of x-axis in anticlockwise direction between 0 and $2\pi$

### Parametric Form

The equation of a straight line passing through the point $(x_1 ,y_1)$ and making an angle $\theta$ with the positive direction of x-axis is

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

Here $0 \leq \theta \leq \pi$

r is the distance of the point (x,y) from $(x_1,y_2)$

The coordinates of any point on the line at a distance r from the point $(x_1, y_1)$ can be taken as
$(x_1 + r cos \theta, y_1 + r sin \theta)$

## Equations of straight lines passing through a given point and making an given angle with a given line

Equation of line passing through the point $(x_1,y_1)$ and making an angle $\alpha$ with the given line $y=m_1 x + c$ is given as

$y-y_1= \frac {m_1 \pm tan \alpha}{1 \mp m_1 tan \alpha} (x-x_1)$

## Length of Perpendicular Point From a Line

The distance of a point from a line is the length of the perpendicular drawn from the point to the line. Let L : Ax + By + C = 0 be a line, whose distance from the point P (x1, y1) is d. Then d is given by

$d= \frac {|Ax_1 + By+1 + C|}{\sqrt {A^2 + B^2}}$

## Distance between two Parallel Lines

For two parallel lines
$y=mx + c_1$
$y=mx+ c_2$
Distance between them is given by
$d= \frac {|c_1 -c_2|}{\sqrt {1+ m^2}}$

## Position of two points with respect to given line

For points $(x_1,y_1)$ and $(x_2,y_2)$ with respect to given line $ax+ By + c =0$

if $\frac {ax_1 + by_1 + c}{ax_2 + by_2 + c} < 0$

Then points are opposite side of the line

if $\frac {ax_1 + by_1 + c}{ax_2 + by_2 + c} > 0$

Then the points are on same side

## Family of Lines passing through intersection of two lines

if $L=0$ and $L^{‘}=0$ be the two line, then family of lines passing through the intersection of two lines is

$L+ \lambda L^{‘}=0$

## Concurrency of Straight lines

For the three lines to be concurrent

$a_1 x + b_1 y + c_1=0$
$a_2 x + b_2 y + c_2=0$
$a_3 x + b_13y + c_3=0$

we should have
$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$

or

$mL_1 + nL_2+ kl_3=0$ where m,n and k are constant which are non zero at the same time

## Equations of the Bisectors of the angles between the lines

The equations of the bisectors of the angles between the lines $a_1 x + b_1 y + c_1=0$ and $a_2 x + b_2 y + c_2=0$ is

$\frac {a_1 x + b_1 y +c_1}{\sqrt {a_1^2 + b_1^2} }=\pm \frac {a_2 x + b_2 y +c_2}{\sqrt {a_2^2 + b_2^2} } How to Find the equation of bisector of obtuse and acute angle Let the line be such that$a_1 x + b_1 y + c_1=0$and$a_2 x + b_2 y + c_2=0$and$c_1 >0$and$c_2> 0$Then the equation of bisector containing Origin$\frac {a_1 x + b_1 y +c_1}{\sqrt {a_1^2 + b_1^2} }= \frac {a_2 x + b_2 y +c_2}{\sqrt {a_2^2 + b_2^2} }$And the other line is$\frac {a_1 x + b_1 y +c_1}{\sqrt {a_1^2 + b_1^2} }= – \frac {a_2 x + b_2 y +c_2}{\sqrt {a_2^2 + b_2^2} }$Now If$a_1a_2 + b_1b_2 >0$then the origin lies in obtuse angle i.e., the bisector containing origin is obtuse angle bisector and if$a_1a_2 + b_1b_2 <0$then the origin lies in acute angle i.e., the bisector containing origin is acute angle bisector. ## Foot of The Perpendicular from a point on Line The co-ordinate of foot of perpendicular$x_2 ,y_2$of a given point$x_1,y_1$on the line$ax+ by+ c=0$are given by$\frac {x_2 -x_1}{a} = \frac {y_2 -y_1}{b} = – \frac {ax_1 + by_1 + c}{a^2 + b^2}$## Image of a point on a Line The co-ordinate of image$x_2 ,y_2$of a given point$x_1,y_1$on the line$ax+ by+ c=0$are given by$\frac {x_2 -x_1}{a} = \frac {y_2 -y_1}{b} = -2 \frac {ax_1 + by_1 + c}{a^2 + b^2}$## Standard Points of the Triangle Centroid Of the Triangle • It is the concurrent point of the Median of the triangle • Centroid divides the median in the ratio 2: 1 • Coordinates are given by$x= \frac {x_1 + x_2 + x_3}{3}$and$y= \frac {y_1+ y_2 + y_3}{3}$Incenter of the Triangle • It is the concurrent point of the angle bisectors • Coordinates are given by$x=\frac {ax_1 + bx_2 + cx_3}{a+ b+ c}$and$y=\frac {ay_1 + by_2 + cy_3}{a+ b+c}$Here a, b, c are the sides of the triangles Circumcenter of the Triangle • It is the concurrent point of the perpendicular bisectors of the sides of the triangles. • If A, B, C are the angles of the$\Delta ABC$and Vertices are $A(x_1, y_1)$ , $B(x_2, y_2)$ and (\ C(x_3, y_3) \), then circumcenter point is given as$x=\frac {x_1 sin2A + x_2 sin2B + x_3 sin2C}{sin2A + sin2B+ sin2C}$and$y=\frac {y_1 sin2A + y_2 sin2B+ y_3 sin2C}{sin2A + sin2B+ sin2C}$Orthocenter of the Triangle • The orthocenter of a triangle is the point of intersection of altitudes • For an acute triangle, it lies inside the triangle. • For an obtuse triangle, it lies outside of the triangle. • If A, B, C are the angles of the$\Delta ABC$and Vertices are $A(x_1, y_1)$ , $B(x_2, y_2)$ and (\ C(x_3, y_3) \), then orthocenter point is given as$x=\frac {x_1 Tan A + x_2 Tan B + x_3 Tan C}{Tan A + Tan B+ Tan C}$and$y=\frac {y_1 Tan A + y_2 Tan B+ y_3 Tan C}{Tan A + Tan B+ Tan C}\$

Important Points

• In right angled the right angled vertex is the orthocenter of triangle and mid-point of hypotenuse is circumcenter
• In a equilateral Triangle, Centroid, Incenter, Circumcenter and Orthocenter coincides

This site uses Akismet to reduce spam. Learn how your comment data is processed.