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

<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line Lecture 4</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Distance of a point from a line</B></Primary></p>
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<p>$AB=|\frac{c}{ab}|\sqrt{a^2+b^2}$</p>
<p>$let PN=d$</p>
<p>$\therefore \text { Area } \triangle P A B  =\frac{1}{2} \times A B \times P N $</p>
<p>$ =\frac{1}{2} \times\left|\frac{c}{a b}\right| \sqrt{a^2+b^2} \times d -\text { (i) }$</p>
<p>area of $\triangle P A B$</p>
<p>$=\frac{1}{2}|x_1(0+\frac{c}{b})+(-\frac{c}{a})(-\frac{c}{b}-y_1)$</p>
<p>$ + 0(y_1-0)|=\frac{1}{2}|\frac{c x_1}{b}+\frac{c^2}{a b}+\frac{c y_1}{a^{\prime}}|$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line Lecture 4</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Distance between two parallel lines</B></Primary></p>
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<p>$a x+b y+c_1=0 \text { - (i) } $</p>
<p>$ a x+b y+c_2=0 \text { - (ii) }$</p>
<p>put $x=0$ in (i)</p>
<p>$ b y+c_1=0 \Rightarrow y=-\frac{c_1}{b} $</p>
<p>$ P\left(0,-\frac{c_1}{b}\right)$</p>
<p>distence of $P$ on (i) from line (ii)</p>
<p>$d=\frac{\left|a \times 0+b \times(-c_ 1 / b)+c_2\right|}{\sqrt{a^2+b^2}}$</p>
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Distance of a point from a line$\to$<span style="color:blue">Distance between two parallel lines</span>$\to$Example$\to$Solution
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line Lecture 4</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Example 3</B></Primary></p>
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<p>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.</p>
<p><strong>Solution:</strong></p>
<p>IN $\triangle ABN, \angle B N A=90^{\circ}$</p>
<p>$\therefore \sin 60^{\circ}=\frac{d}{a} $</p>
<p>$ \Rightarrow \quad \frac{\sqrt{3}}{2}=\frac{d}{a} $</p>
<p>$ \Rightarrow d=\frac{\sqrt{3}}{2} a-(i)$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line Lecture 4</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Solution</B></Primary></p>
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<p>$d  =\frac{\sqrt{3}}{2} a \quad $ from (i)</p>
<p>$\Rightarrow 4 \sqrt{2}  =\frac{\sqrt{3}}{2} a $</p>
<p>$a  =\frac{8 \sqrt{2}}{\sqrt{3}}$</p>
<p>$\therefore$ Area of $\triangle A B C=\frac{\sqrt{3}}{4} a^2$</p>
<p>$ =\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}  $</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line Lecture 4</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Solution</B></Primary></p>
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<p>$d  =\frac{|-1-1-6|}{\sqrt{(+1)^2+(+1)^2}} $
$ =\frac{|-8|}{\sqrt{2}}=\frac{8}{\sqrt{2}}=\frac{8 \sqrt{2}}{2}$</p>
<p>In $\triangle A B N$</p>
<p>$ \sin 60^{\circ}=\frac{d}{a} $</p>
<p>$ \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}}$</p>
<p>$\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}$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line Lecture 4</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Example 5</B></Primary></p>
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<p>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
$</p>
<p><strong>Solution:</strong></p>
<p>Given lines $3 x+2 y+6=0$  $\rightarrow$ (i)</p>
<p>(i) $9 x+6 y-7=0$</p>
<p>(ii) $\Rightarrow 3\left(3 x+2 y-\frac{7}{3}\right)=0$ $\rightarrow$ (ii)</p>
<p>$3 x+2 y+k=0$, this line is (ii) to line (i)</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line Lecture 4</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Solution</B></Primary></p>
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<p>$
3 x+2 y+k=0
$</p>
<p>put $x=0, y=-\frac{k}{2}$</p>
<p>$
P(0,-k / 2)
$</p>
<p>according to question $\quad d_1 = d_2$</p>
<p>$ \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}} $</p>
<p>$ \Rightarrow \quad |-k+6|=|-k-7 / 3|=| k+7 / 3 \mid $</p>
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<p>$ \Rightarrow \quad-k+6= \pm(k+7 / 3)$</p>
<p>$\Rightarrow ~ -k+6=k+7 / 3$</p>
<p>or $-k+6=-k-7 / 3 $</p>
<p>$\Rightarrow-2 k=-6+7 / 3 $</p>
<p>$ \Rightarrow-2 k=-\frac{18+7}{3} \Rightarrow \quad+2 k= - \frac{11}{3}$</p>
<p>$ \Rightarrow k=\frac{11}{6}$</p>
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<p>======</p>
<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line Lecture 4</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Solution</B></Primary></p>
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<p>$\therefore ~ $ equation of line, $3 x+2 y+k=0$</p>
<p>$\Rightarrow 3 x+2 y+ \frac {11} {6}=0$</p>
<p>$\Rightarrow 18 x+12 y+11=0$</p>
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<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Straight line Lecture 4</B></Success></p>
<p><Primary style="float:center;text-align:center;font-size:24px;"><B>Example 6</B></Primary></p>
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<p>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 $(-4,1)$.</p>
<p><strong>Solution:</strong></p>
<p>Given equation It line $12 x+5 y=1)$ - (i) equation of line perpendicular to line (i) $5 x-12 y+k=0$</p>
<p>$  \frac{|5 \times(-4)+| 2 \times 1+k \mid}{\sqrt{5^2+12^2}}=2 $</p>
<p>$\Rightarrow \frac{|-20+12+k|}{\sqrt{169}}=2 $</p>
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<p>$\rightarrow 3 y-9=5 x-10$</p>
<p>$\rightarrow \quad 5 x-3 y-1=0$</p>
<p>Equation of BC</p>
<p>$(y+1)=-\frac{3}{5}(x-4)$</p>
<p>$\Rightarrow 5 y+5=-3 x+12 $
$\Rightarrow  3 x+5 y-7=0 $</p>
<p>$\therefore ~ A  =\frac{\mid 3 \times 2+5 \times 3 -7\mid}{\sqrt{3^2+5^2}} $
$=\frac{|14|}{\sqrt{34}}=\frac{14}{\sqrt{34}}$</p>
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