mathematics-class-11-unit-11-chapter-2-straight-lines-l-2-6-wncn38qokzg

Straight line lecture 2

Parallel and perpendicular lines

1. Parallel lines

$l_1 ~ \text {parallel} ~ l_2$

$\Rightarrow \theta_1=\theta_2$

$\Rightarrow \tan \theta_1=\tan \theta_2$

$\Rightarrow m_1=m_2$

$(l_1) \text {parallel} (l_2) \Longleftrightarrow m_1=m_2$

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Straight line lecture 2

Parallel and perpendicular lines

2. Perpendicular Lines

$l_1 \perp l_2$

$\theta_1 =90^{\circ}+\theta_2$ $\Rightarrow \tan \theta_1 =\tan(90^{\circ}+\theta_2)$

$\Rightarrow m_1 =-\cot \theta_2$ $=-\frac{1}{\tan \theta}$

$\Rightarrow m_1=-\frac{1}{m}$ $\Rightarrow m_ 1 \times m_2=-1$

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Straight line lecture 2

Example 1

Show that the line joining $(2,-3)$ and $(-5,1)$ is

(i) parallel to the line joining $(7,-1)$ and $(0,3)$

(ii) perpendicular to the line joining $(4,5)$ and $(0,-2)$

Straight line lecture 2

Solution

$\quad P(2,-3) \text{and} Q(-5,1)$

$\therefore \text{slope of} \ P Q =\frac{1+3}{-5-2}=\frac{4}{-7}$

(i) $A(7,-1)$ and $B(0,3)$

$\therefore$ slope of $A B=\frac{3+1}{0-7}=\frac{4}{-7}$

$\therefore$ slope of $P Q=$ slope of $A B=-\frac{4}{7}$

$\therefore \quad P Q || A B$

Straight line lecture 2

Solution

(ii)

$C(4,5)$ and $D(0,-2)$

$\therefore$ Slope of $C D=\frac{-2-5}{0-4}$

$=\frac{-7}{-4}=\frac{7}{4} (m_1)$

Now, Slope of $P Q=\frac{4}{-7}(m_2)$

$m_1 \times m_2=\frac{7}{4} \times \frac{4}{-7}=-1$

$\Rightarrow C D \perp P Q .$

Straight line lecture 2

Angle between two lines

Let angle with $x$ -axis by the line $l_1$ and $l_2$ and $\theta_1$ and $\theta_2$ respectively.

$\therefore$ slope of

$l_1(m_1)=\tan \theta_1$

slope of $l_2(m_2) =\tan \theta_2$

$\theta=\theta_2-\theta_1$

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Straight line lecture 2

Angle between two lines

$\therefore \tan \theta =\tan(\theta_2-\theta_1)$

$=\frac{\tan \theta_2-\tan \theta_1}{1+\tan \theta_2 \cdot \tan \theta_1}$

$\therefore \tan \theta=\frac{m_2-m_1}{1+m_2 m_1}$

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

$= \pm \frac{m_2-m_1}{1+m_1 m_2}$

$+\rightarrow$ acute angle between $l_1$ and $l_2$

$-\rightarrow \text {obtuse angle between} l_1 \text{and} l_2$

Straight line lecture 2

Example 2

Find the angle between the St. lines whose slopes are $-\frac{7}{3}$ and $\frac{5}{2}$ .

Solution:

Given $m_1=-\frac{7}{3}$ and $m=\frac{5}{2}$

$\therefore {Let,}\quad \theta$ be the angle between the lines

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

$=|\frac{5 / 2+7 / 3}{1+5 / 2 \times(-7 / 3)}|$ $=|\frac{29 / 6}{-29 / 6}|=|-1|$

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Straight line lecture 2

Example 3

If the angle between two lines is $\pi / 4$ and the slope of one of the lines is $1 / 2$ . Find the slope of the other line.

Solution:

Given,

$\theta=\pi / 4$ and $m_1=\frac{1}{2}$ , then $m_2=$ ?

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

$\Rightarrow \tan \pi / 4=\frac{m_2-\frac{1}{2}}{1+\frac{m_2}{2}}$

$\Rightarrow 1=\frac{(2 m_2-1) / 2}{(2+m_2) / 2}$

Straight line lecture 2

Solution

$\Rightarrow \quad \frac{2+m_2}{2}$

$=\frac{2 m_2-1}{2}$

$\Rightarrow 2 m_2-m_2=2+1$

$\Rightarrow m_2=3$

$\therefore$ slope of $2^{nd}$ line is 3 .

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Straight line lecture 2

Equation of a straight line

The equation of a line is an equation in $x, y$ which is satisfied by every point of the line.

Straight line lecture 2

Equation of a line parallel to x-axis

Equation of line parallel to $x$ -axis is $y=b$

Example:

$y=1, y=-2$

$y=13 / 5$

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Straight line lecture 2

Equation of a line parallel to y-axis

Equation of $l$ parallel to $y$

axis is $x=a$

Example: $x=-1, x=7$ ,

$x=-\frac{1}{2}$

Straight line lecture 2

Example 4

Find the equation of the line passing through $(2,3)$ and is

(i) parallel to $x$ -axis

(ii) parallel to $y$ -axis.

Straight line lecture 2

Solution

$y =3$

$\therefore \quad$ Equation of $l_1$ is y = 3

$x=2$

Equation of $l_2$ is $x=2$

Straight line lecture 2

Equation of straight line in various standard forms

1. Point- Slope Form:

Let line passing through $P(x_1,y_1)$

with slope $m$ .

$m=\tan \theta=\frac{Q R}{P R}=\frac{y-y_1}{x-x_1}$

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

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Straight line lecture 2

Equation of straight line: slope-intercept form

Let line having slope $m$ . and $y$ -intercept (c).

$\Rightarrow$ line $l$ passing through $Q(0, c)$

$\therefore$ Equation of line

$(y-c)=m(x-0)$

$\Rightarrow y-c=m x$

$\Rightarrow y=m x+c$

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Straight line lecture 2

Slope - intercept form

$y=m x+c$

(i) When $m \neq 0$ and $c=0 \Rightarrow y=m x$ $\rightarrow$ line passing through origion.

(ii) When $m=0$ and $c=0 \Rightarrow y=0$ $\Rightarrow$ line coincides with $x$ -axis.

(iii) When $m=0$ and $c \neq 0$

$\Rightarrow y=c$

$\Rightarrow$ line parallel to $x$ -axis

Straight line lecture 2

Equation of straight line: two point form

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

Let line passing through $P(x_1, y_1)$ and slope $m$ ,

then, equation of line

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

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

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

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Straight line lecture 2

Equation of straight line: intercept form

Let line $l$ makes $x$ -intercept and $y$ -intercept $a$ and $b$ respectively.

is line passing through $A(a,0)$ and $B(0,b)$

$\therefore \text{slope of} A B(m)=\frac{b-0}{0-a}=-\frac{b}{a}$

$\therefore$ Equation of line through $A(a,0)$ and slope

$=-b / a$ is $(y-0) =-\frac{b}{a}(x-a)$

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

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Straight line lecture 2

Equation of straight line: perpendicular or normal form

$A(p \cos \alpha, p \sin \alpha)$

$\therefore \quad$ slope of OA $=\tan \alpha$ Since OA is perpendicular to $l$ (given)

$\Rightarrow$ slope of $l$ ie. $m=-\frac{1}{\tan \alpha} = -\cot \alpha$

Equation of $l$ passing through

$A(p \cos \alpha, p \sin \alpha)\text {with slope}$

$m=-\cot \alpha$ is

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Straight line lecture 2

Equation of straight line: perpendicular or normal form

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

$\Rightarrow (y-p \sin \alpha)=m(x-p \cos \alpha)$

$\Rightarrow (y-p \sin \alpha) =-\frac{\cos \alpha}{-\sin \alpha}(x-p \cos \alpha)$

$\Rightarrow y \sin \alpha-p \sin ^2 \alpha=-x \cos\alpha+p \cos ^2 \alpha$

$\Rightarrow x \cos \alpha+y \sin \alpha =p \sin ^2 \alpha+p \cos ^2 \alpha$

$\quad \quad =p(\sin ^2 \alpha+\cos ^2 \alpha) =p$

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

Straight line lecture 2

Equation of straight line: general form

$A x+B y+C=0, \quad$ $A,B$ and $C \in R .$

$A$ and $B$ not both equal to zero.

${A=0} \Rightarrow B y+C=0 \Rightarrow y=-C / B$
$B=0 \Rightarrow A x+C=0 \Rightarrow x=-C / A$ and $A \neq 0$
$A \neq 0$ and $B \neq 0$

Straight line lecture 2

Equation of straight line: general form

$A x+B y+c=0$

$\Rightarrow \quad B y=-A x-C$

$\Rightarrow \quad y=-\frac{A}{B} x-\frac{C}{B}$

Thank you