mathematics-class-11-unit-11-chapter-4-straight-lines-l-4-6-7e-6uoy6gky

Straight line Lecture 4

Distance of a point from a line

$a x+b y+c=0$

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

$\therefore O A=-c / a$

$O B=-c / b$

$A B=\sqrt{\left(-\frac{c}{a}+0\right)^2+\left(0+\frac{c}{a}\right)^2}$

$=\sqrt{\frac{c^2}{a^2}+\frac{c^2}{b^2}}$

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Straight line Lecture 4

Distance of a point from a line

$AB=|\frac{c}{ab}|\sqrt{a^2+b^2}$

$let PN=d$

$\therefore \text { Area } \triangle P A B =\frac{1}{2} \times A B \times P N$

$=\frac{1}{2} \times\left|\frac{c}{a b}\right| \sqrt{a^2+b^2} \times d -\text { (i) }$

area of $\triangle P A B$

$=\frac{1}{2}|x_1(0+\frac{c}{b})+(-\frac{c}{a})(-\frac{c}{b}-y_1)$

$+ 0(y_1-0)|=\frac{1}{2}|\frac{c x_1}{b}+\frac{c^2}{a b}+\frac{c y_1}{a^{\prime}}|$

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Straight line Lecture 4

Distance of a point from a line

area: $\triangle$ (P A B)

$=\frac{1}{2}\left|\frac{c}{a b}\right|\left|a x_1+b y_1+c\right|-\text { (ii) }$

from (i) & (ii)

$\frac{1}{2} \times|\frac{c}{a b}|\sqrt{a^2+b^2} \times d$

$=\frac{1}{2}| \frac{c}{a b}|| a x_1+b y_1+c \mid.$

$\therefore d=\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}$

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Straight line Lecture 4

Distance between two parallel lines

$a x+b y+c_1=0 \text { - (i) }$

$a x+b y+c_2=0 \text { - (ii) }$

put $x=0$ in (i)

$b y+c_1=0 \Rightarrow y=-\frac{c_1}{b}$

$P\left(0,-\frac{c_1}{b}\right)$

distence of $P$ on (i) from line (ii)

$d=\frac{\left|a \times 0+b \times(-c_ 1 / b)+c_2\right|}{\sqrt{a^2+b^2}}$

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Straight line Lecture 4

Distance between two parallel lines

$\therefore d=\frac{\left|-c_2+c_1\right|}{\sqrt{a^2+b^2}}=\frac{\left|c_2-c_1\right|}{\sqrt{a^2+b^2}}$

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Straight line Lecture 4

Example 1

Find the distance of the point $(-2,3)$ from the line $12 x-5 y=2$ .

Solution:

$12 x-5 y=2$

put $x=0 \Rightarrow y=\frac{-2}{5}$

put $y=0 \Rightarrow x=\frac{1}{6}$

$d=\frac{\left|a x_ 1+b y_1+c\right|}{\sqrt{a^2+b^2}}$

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Straight line Lecture 4

Solution

Here

$a=12, b=-5, c=-2$

$x_1=-2$ , $y_1=3$

$d =\frac{|12 \times(-2)+(-5) \times 3-2|}{\sqrt{(12)^2+(-5)^2}}$

$=\frac{|-24-15-2|}{\sqrt{169}}=\frac{41}{13} \text { units. }$

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Straight line Lecture 4

Example 2

Find the distance between the lines $3 x+4 y=9$ and $6 x+8 y=15$ .

Solution:

Given lines $3 x+4 y=9$

$\Rightarrow$ $3 x+4 y-9=0$

$6 x+8 y=15$ $\Rightarrow$ $2(3 x+4 y)=15$

$\Rightarrow$ $3 x+4 y-$ $\frac{15}{2}=0$ $\quad \because (\frac{-15}{2}=c_2)$

$d=\frac{\left|c_2-c_1\right|}{\sqrt{a^2+b^2}}=\frac{\left|-\frac{15}{2}+9\right|}{\sqrt{3^2+4^2}}$

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Straight line Lecture 4

Solution

$d=\frac{\left|\frac{-15+18}{2}\right|}{\sqrt{25}}=\frac{3 / 2}{5}=\frac{3}{10}$ units.

Straight line Lecture 4

Example 3

If the equation of the base of an equilateral traingle is $x+y-6=0$ and the opposite vertex is the point $(-1,-1)$ , then find the area of the traingle.

Solution:

IN $\triangle ABN, \angle B N A=90^{\circ}$

$\therefore \sin 60^{\circ}=\frac{d}{a}$

$\Rightarrow \quad \frac{\sqrt{3}}{2}=\frac{d}{a}$

$\Rightarrow d=\frac{\sqrt{3}}{2} a-(i)$

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Straight line Lecture 4

Solution

Slope of $B C=-1$

Slope of $A N=1$ (AN perpendicular $B C$ )

equation of AN, $(y+1)=1(x+1)$

$\Rightarrow$ $x-y=0$ - (ii)

$x+y-6=0$ -(iii)

Straight line Lecture 4

Solution

from (ii) & (iii)

$x+x=6 \Rightarrow x=3$

$\quad \therefore y=3$

$A N= \sqrt{(3+1)^2+(3+1)^2}= \sqrt{4^2+4^2}=4 \sqrt{2}$

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Straight line Lecture 4

Solution

$d =\frac{\sqrt{3}}{2} a \quad$ from (i)

$\Rightarrow 4 \sqrt{2} =\frac{\sqrt{3}}{2} a$

$a =\frac{8 \sqrt{2}}{\sqrt{3}}$

$\therefore$ Area of $\triangle A B C=\frac{\sqrt{3}}{4} a^2$

$=\frac{\sqrt{3}}{4}\left(\frac{8 \sqrt{2}}{\sqrt{3}}\right)^2$ $=\frac{\sqrt{3}}{4} \times \frac{64 x^2}{3}$ $=\frac{32 \sqrt{3}}{3}$

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Straight line Lecture 4

Solution

$d =\frac{|-1-1-6|}{\sqrt{(+1)^2+(+1)^2}}$ $=\frac{|-8|}{\sqrt{2}}=\frac{8}{\sqrt{2}}=\frac{8 \sqrt{2}}{2}$

In $\triangle A B N$

$\sin 60^{\circ}=\frac{d}{a}$

$\Rightarrow \quad \frac{\sqrt{3}}{2}= \frac {\frac{8 \sqrt{2}}{2}}{a}$ $\Rightarrow a \sqrt{3}=a \sqrt{2} \Rightarrow a=\frac{8 \sqrt{2}}{\sqrt{3}}$

$\begin{aligned} \operatorname{ar} & (\triangle A B C) \ & =\frac{1}{2} \times a \times d \ = & \frac{1}{2} \times \frac{8 \sqrt{2}}{\sqrt{3}} \times \frac{8 \sqrt{2}}{2} =\frac{16 {\sqrt{2}}}{\sqrt3} \end{aligned}$

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Straight line Lecture 4

Example 4

If $p . q$ are the lengths of perpendculars from the origin to the lines $x \cos \theta-y \sin \theta=k \cos 2 \theta$ and $x \sec \theta+y \operatorname{cosec} \theta=k$ respectively.

Prove that $p^2+4 q^2=k^2$

$x \cos \theta-y \sin \theta=k \cos 2 \theta$

$\Rightarrow x \cos \theta-y \sin \theta-k \cos 2 \theta=0$

$P$ is the distance of line (i) from origin $(A / Q)$

Straight line Lecture 4

Solution

$p =\frac{|0-0-k \cos 2 \theta|}{\sqrt{\cos ^2 \theta+(-\sin \theta)^2}}$

$=\frac{k \cos 2 \theta}{\sqrt{\cos ^2 \theta+\sin ^2 \theta}}=k \cos 2 \theta$

Again,

$x \sec \theta+y \operatorname{csec} \theta=k$

$\Rightarrow \frac{x}{\cos \theta}+\frac{y}{\sin \theta}=k$

$\Rightarrow x \sin \theta+y \cos \theta=k \sin \theta \cdot \cos \theta$

Straight line Lecture 4

Solution

$\Rightarrow x \sin \theta+y \cos \theta=\frac{1}{2} k \times 2 \sin \theta \cdot \cos \theta$

$=\frac{k}{2} \sin 2 \theta$

$\Rightarrow x \sin \theta+y \cos \theta-\frac{k}{2} \sin 2 \theta=0$

$q=\frac{|0+0-\frac{k}{2} \sin 2 \theta \mid}{\sqrt{\sin ^2 \theta+\cos ^2 \theta}}$

$= \frac{k}{2} \sin ^2 \theta$

Straight line Lecture 4

Solution

$p^2+4 q^2 =(k \cos 2 \theta)^2+4 \times\left(\frac{k}{2} \sin 2 \theta\right)^2$

$=k^2 \cos ^2 2 \theta+4 k \frac{k^2}{x} \sin ^2 2 \theta$

$=k^2\left(\cos ^2 2 \theta+\sin ^2 2 \theta\right)$

$=k^2$

Straight line Lecture 4

Example 5

Find the equation of the line Which is equidistant from the parallel line $3 x+2 y+6=0$ and $9 x+6 y-7=0$

Solution:

Given lines $3 x+2 y+6=0$ $\rightarrow$ (i)

(i) $9 x+6 y-7=0$

(ii) $\Rightarrow 3\left(3 x+2 y-\frac{7}{3}\right)=0$ $\rightarrow$ (ii)

$3 x+2 y+k=0$ , this line is (ii) to line (i)

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Straight line Lecture 4

Solution

$3 x+2 y+k=0$

put $x=0, y=-\frac{k}{2}$

$P(0,-k / 2)$

according to question $\quad d_1 = d_2$

$\Rightarrow \frac{\left|3 \times 0+2 \times\left(-\frac{k}{2}\right)+6\right|}{\sqrt{3^2+2^2}}$ $=\frac{\left|3 \times 0+2 \times(-k / 2)-\frac{7}{3}\right|}{\sqrt{3^2+2^2}}$

$\Rightarrow \quad |-k+6|=|-k-7 / 3|=| k+7 / 3 \mid$

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Straight line Lecture 4

Solution

$\Rightarrow \quad-k+6= \pm(k+7 / 3)$

$\Rightarrow ~ -k+6=k+7 / 3$

or $-k+6=-k-7 / 3$

$\Rightarrow-2 k=-6+7 / 3$

$\Rightarrow-2 k=-\frac{18+7}{3} \Rightarrow \quad+2 k= - \frac{11}{3}$

$\Rightarrow k=\frac{11}{6}$

Straight line Lecture 4

Solution

$\therefore ~$ equation of line, $3 x+2 y+k=0$

$\Rightarrow 3 x+2 y+ \frac {11} {6}=0$

$\Rightarrow 18 x+12 y+11=0$

Straight line Lecture 4

Example 6

Find the equation of stright lines Which are perpendicular to the line $12 x+5 y=17$ and a distance of 2 unit from the point $(-4,1)$ .

Solution:

Given equation It line $12 x+5 y=1)$ - (i) equation of line perpendicular to line (i) $5 x-12 y+k=0$

$\frac{|5 \times(-4)+| 2 \times 1+k \mid}{\sqrt{5^2+12^2}}=2$

$\Rightarrow \frac{|-20+12+k|}{\sqrt{169}}=2$

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Straight line Lecture 4

Solution

$\Rightarrow \quad \frac{|-8+k|}{13}=2$

$\Rightarrow \quad|-8+k|=26$

$\Rightarrow \quad-8+k= \pm 26 \Rightarrow \quad k=8 \pm 26$

$\Rightarrow \quad k=34,-18$

Equation of required line will be

$5 x-12 y+3 y=0$

$5 x-12 y-18=0$

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Straight line Lecture 4

Example 7

In the traingle with vertex $A(2,3), B(4,-1)$ and $C(-1,2)$ , find the equation and length of altitude from the vertex $A$ .

Solution:

slope of $BC=\frac{2+1}{-1-4}$

$=-\frac{3}{5}$

slope of A N $=\frac{5}{3}$

Equation of A N, $(y-3)=\frac{5}{3}(x-2)$

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Straight line Lecture 4

Solution

$\rightarrow 3 y-9=5 x-10$

$\rightarrow \quad 5 x-3 y-1=0$

Equation of BC

$(y+1)=-\frac{3}{5}(x-4)$

$\Rightarrow 5 y+5=-3 x+12$ $\Rightarrow 3 x+5 y-7=0$

$\therefore ~ A =\frac{\mid 3 \times 2+5 \times 3 -7\mid}{\sqrt{3^2+5^2}}$ $=\frac{|14|}{\sqrt{34}}=\frac{14}{\sqrt{34}}$

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Thank you