### <Success>Straight line lecture 6</Success> <div style='float: left; width:100%;font-size:30px;'> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/013-0027.6.jpg"></p> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/57822196-60d4-4e1f-ad68-a00a916c7700/public"></p> </div>
### <Success>Problem 9</Success> <div style='float: left; width:70%;font-size:28px;'> <p>Two vertices of a traingle are $B(5,-1)$ and $c(-2,3)$. If the orthocentre of the traingle is at origin, find the third vertex.</p> <p><strong>Solution:</strong> Slope of $ OB =\frac{0+1}{0-5}=-\frac{1}{5} $</p> <p>slope $ B E=-\frac{1}{5}$</p> <p>$\therefore \text { Slope f } A C=\frac{-1}{-1 / 5}=5 $</p> <p>as, $\quad B E \perp\text { AC }$</p> </div> <div style='float: right; width:30%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/337f2dd1-2612-45c9-eb95-3a2639fc5400/public"></p> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/fa2224e2-ecb9-4a08-512d-574b8b65a500/public"></p> <!----> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Slope of $C F=\frac{0-3}{0+2}=-\frac{3}{2}$</p> <p>Since $C F \perp A B$</p> <p>$\Rightarrow \text { slope of } A B=\frac{-1}{-3/2}=2 / 3$</p> <p>Equation of $A C$ with slope $=5 \rightarrow$ passing through $c(-2,3)$</p> <p>$y-3=5(x+2)$</p> <p>$\Rightarrow \quad 5 x-y+13=0 \cdots (1)$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/000001.png"></p> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Again equation of line $A B$ with slope: $2 / 3$ and passing through $B(5,-1)$</p> <p>$ y+1=\frac{2}{3}(x-5) $</p> <p>$\Rightarrow 3 y+3=2 x-10$</p> <p>$\Rightarrow 2 x-3 y-13=0 \cdots(2)$</p> <p>from (1) and (2)</p> <p>$5 x-y+13=0$</p> <p>$\Rightarrow y=5 x+13$</p> <p>put $y=5 x+13$ in (2)</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/083-0564.0.jpg"></p> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$2 x-3(5 x+13)-13=0$</p> <p>$\Rightarrow-13 x-52=0$</p> <p>$\Rightarrow \quad x=-4 $</p> <p>$ \therefore y=5 x+13$</p> <p>$=5 x(-4)+13=-7$</p> <p>$\therefore$ This vertex $A(-4,-7)$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/091-0637.2.jpg"></p> </div>
### <Success>Problem 10</Success> <div style='float: left; width:60%;font-size:28px;'> <p>A straight line is drawn through the point $P(1,0)$. It intersects the straight line $y=2 x-3$ at the point $Q$. Find the slopes of the straight line $P Q$ if $P Q=\sqrt{2}$.</p> <p><strong>Solution:</strong></p> <p>Given equation of line $ y=2 x-3 $</p> <p>Let $Q(\alpha, \beta)$</p> <p>Q lies on $y=2 x-3$</p> <p>$\therefore \beta=2 \alpha-3 $</p> <p>$\therefore \quad Q(\alpha, 2 \alpha-3) $</p> <p>$A / q \quad P Q=\sqrt{2}$</p> </div> <div style='float: right; width:40%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/1105f9ad-834e-48af-0910-27ebde6ff700/public"></p> <!----> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ \Rightarrow \quad P Q^2=2 $</p> <p>$\Rightarrow \quad(\alpha-1)^2+(2 \alpha-3)^2=2$</p> <p>$ \Rightarrow \quad \alpha^2-2 \alpha+1+4 \alpha^2$</p> <p>$=12 \alpha+9=2$</p> <p>$ \Rightarrow \quad 5 \alpha^2-14 \alpha+10=2$</p> <p>$ \Rightarrow \quad 5 \alpha^2-14 \alpha+8=0$</p> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ \Rightarrow 5 \alpha^2-10 \alpha-4 \alpha+8=0 $</p> <p>$ \Rightarrow \quad 5 \alpha(\alpha-2)-4(\alpha-2)=0 $</p> <p>$\Rightarrow \quad(5 \alpha-4)(\alpha-2)=0 $</p> <p>$\Rightarrow \quad \alpha=\frac{4}{5}, 2$</p> <p>$\therefore \beta =2 \alpha-3$</p> <p>$\Rightarrow \beta =2 \times \frac{4}{5}-3$</p> <p>or $\quad\beta=2 \times 2-3 $</p> <p>$ \beta=\frac{8-15}{5} $</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/000002.png"></p> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\Rightarrow \beta =-\frac{7}{5}$</p> <p>or $ \beta=1 $</p> <p>$\therefore Q = (\frac{4}{5},-\frac{7}{5})$</p> <p>$Q^{\prime}=(2,1)$</p> <p>Given $P(1,0)$</p> <p>$\therefore \text { slope of } P Q=\frac{-7 / 5-0}{4 / 5-1}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/140-1056.0.jpg"></p> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ =\frac{-7 / 5}{-1 / 5}=7 $</p> <p>Slope of $P Q^{\prime}=\frac{1-0}{2-1}=\frac{1}{1}=1$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/145-1094.4.jpg"></p> </div>
### <Success>Problem 11</Success> <div style='float: left; width:60%;font-size:29px;'> <p>Show that the four lines $a x \pm b y \pm c=0$ enclose a rhombus whose area is $2 c^2 / a b$.</p> <p><strong>Solution:</strong></p> <p>Given equation</p> <p>$a x \pm b y \pm c =0 $</p> <p>i.e. $\ a x+b y+c =0-(i) $</p> <p>$a x+b y-c =0-(ii) $</p> <p>$a x-b y+c =0 - (iii) $</p> <p>$a x-b y-c =0 - (iv) $</p> </div> <div style='float: right; width:40%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/0ef3f343-eab1-4777-2ae4-8409e64b8500/public"></p> <!----> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ a x+b y+c=0 $</p> <p>$\Rightarrow \frac{x}{-c / a}+\frac{y}{-c / b}=1$</p> <p>$ a x+b y-c=0 $</p> <p>$\Rightarrow \frac{x}{c / a}+\frac{y}{c / b}=1 $</p> <p>$ a x-b y+c=0 $</p> <p>$\Rightarrow \frac{x}{-c / a}+\frac{y}{c / b}=1 $</p> <p>$ a x-b y-c=0 $</p> <p>$\Rightarrow \frac{x}{c / a}+\frac{y}{-c / b}=1$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/165-1322.4.jpg"></p> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <!--11. Show that the four lines $a x \pm b y \pm c=0$ enclose a rhombus whose area is $2 c^2 / a b$.--> <!--Given equation--> <!--$a x \pm b y \pm c =0 $--> <!-- That is :--> <!--$a x+b y+c =0-i $4--> <!--$a x+b y-c =0-i $--> <!--$a x-b y+c =0-iii $--> <!--$a x-b y-c =0-iv $--> <p>$ B D=\sqrt{O+(\frac{c}{b}+\frac{c}{b})^2}=\frac{2 c}{b} $</p> <p>$A C=\sqrt{(\frac{c}{a}+\frac{c}{a})^2+0}=\frac{2 c}{a}$</p> <p>$\text { Area of rhombus } =\frac{1}{2} \times d_1 \times d_2 $</p> <p>$=\frac{1}{2} \times B D \times A C $</p> <p>$=\frac{1}{2} \times \frac{2 c}{b} \times \frac{2 c}{a} $</p> <p>$ =\frac{2 c^2}{a b}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/178-1383.6.jpg"></p> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/227e66be-c162-4dcd-7b2d-5416b39a2a00/public"></p> </div>
### <Success>Problem 12</Success> <div style='float: left; width:70%;font-size:28px;'> <p>Two sides of a square lie on the lines $x+y=1$ and $x+y=-2$. Find its area.</p> <p><strong>Solution:</strong></p> <p>Given equation of lines</p> <p>$x+y=1 \cdots(i)$</p> <p>$x+y=-2\cdots(ii)$</p> <p>$\because$ slopes of lines (i) and (ii) are equal that is slope $=-1$</p> <p>$\therefore \quad$ line are parallel distance between parallel lines</p> <p>$=\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}$</p> </div> <div style='float: right; width:30%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/30867750-2327-4882-6583-e6c546831600/public"></p> <!----> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>in the given equation $c_1=-1$ and $c_2=2$</p> <p>$\therefore \text {distance}$</p> <p>$(d) =\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}$</p> <p>$=\frac{|2+1|}{\sqrt{1^2+1^2}} =\frac{3}{\sqrt{2}}$</p> <p>$\therefore$ length of sides of square.</p> <p>$=$ distance between given parallel lines $=\frac{3}{\sqrt{2}}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/000003.png"></p> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\text { Area of square }$</p> <p>$ =d^2=\left(\frac{3}{\sqrt 2}\right)^2 $</p> <p>$=\frac{9}{2} \text { square units. }$</p> <!--12. Two sides of a square lie on the lines $x+y=1$ and $x+y=-2$. Find its area--> <!--Given equation of lines--> <!--$x+y=1 \cdots(i)$--> <!--$x+y=-2\cdots(ii)$--> <!--Since slopes of lines (i) and (ii) are equal that is slope $=-1$--> <!--$\therefore$ line are parallel--> <!--distance between parallel lines =--> <!--$\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}$--> <!--.--> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/0313778e-2509-4461-81f6-f45b5ae15e00/public"></p> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/230-1822.8.jpg"></p> </div>
### <Success>Problem 13</Success> <div style='float: left; width:70%;font-size:29px;'> <p>If the line joining two points $A(2,0)$ and $B(3,1)$ is rotated about $A$ in anti-clockwise direction through an angle of $15^{\circ}$, find the equation of the line in new position.</p> <p><strong>Solution:</strong></p> <p>Let- the line makes angle $\theta$ with $x$-axis.</p> <p>$ \text { slope of line }=\tan \theta $</p> <p>$\Rightarrow \frac{1-0}{3-2}=\tan \theta $</p> <p>$\Rightarrow \tan\theta =1 =\tan 45^{\circ}$</p> <p>$\Rightarrow \ \theta=45^{\circ}$</p> </div> <div style='float: right; width:30%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/1901d99c-ee21-47cf-eb98-1c083c5ff200/public"></p> <!----> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>In new position when line rotates about $A(2,0)$ in anticlockwise then resultant angle with $x$-axis</p> <p>$=\theta+15^{\circ}$</p> <p>$=45^{\circ}+15^{\circ}$</p> <p>$=60^{\circ}$</p> <p>$\therefore$ slope of line in new position.</p> <p>$=\tan 60^{\circ}$</p> <p>$=\sqrt{3}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/260-2085.6.jpg"></p> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\therefore$ Equation of line with slolpe $\sqrt 3$ and passing through the point $A(2,0)$</p> <p>$ y-0=\sqrt{3}(x-2) $</p> <p>$ \Rightarrow \quad y=\sqrt{3} x-2 \sqrt{3} $</p> <p>$\Rightarrow \quad \sqrt{3} x-y-2 \sqrt{3}=0$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/268-2134.8.jpg"></p> </div>
### <Success>Problem 14</Success> <div style='float: left; width:70%;font-size:30px;'> <p>Find the equation of the bisector of $\angle A$ of $\triangle A B C$, whose vertices are $A(4,3), B(0,0)$ and $C(2,3)$</p> <p><strong>Solution:</strong></p> <p>$\frac{A B}{A C}=\frac{B D}{D C}$</p> <p>$A B=\sqrt{4^2+3^2}=\sqrt{25}=5$</p> <p>$A C=\sqrt{(4-2)^2+(3-3)^2}=\sqrt{2^2}=2$</p> <p>$\therefore \frac{B D}{D C}=\frac{A B}{A C}=\frac{5}{2}$</p> </div> <div style='float: right; width:30%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/25bb949e-17a1-4e33-5871-3d26e5602400/public"></p> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/56da8943-5f2e-4079-e0e4-de72e60f1600/public"></p> <!----> </div>
### <Success>Solution</Success> <div style='float: left; width:60%;font-size:30px;'> <p>Point $D(\frac{2 \times 0+5 \times 2}{5+2}, \frac{5 \times 3+2 \times 0}{5+2})$</p> <p>i.e.,</p> <p>$D(\frac{10}{7}, \frac{15}{7})$</p> <p>Equation of $A D$</p> <p>$(y-3) =\frac{\frac{15}{7}-3}{\frac{10}{7}-4}(x-4) $</p> <p>$\Rightarrow(y-3) =\frac{1}{3}(x-4) $</p> <p>$\Rightarrow 3 y-9 =x-y \Rightarrow x-3 y+5=0$</p> </div> <div style='float: right; width:40%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/1d0fe9a1-a11e-490f-3b56-8568010fe800/public"></p> <!----> </div>
### <Success>Problem 15</Success> <div style='float: left; width:65%;font-size:29px;'> <p>A ray of light -passing through the point $P(1,2)$ reflects on $x$-axis at the point $A$ and the reflected ray passes through the point $Q(5,3)$.</p> <p>Find the coordinate of $A$.</p> <p><strong>Solution:</strong></p> <p>Let, reflected ray $AQ$ makes angle $\theta$ with the $x$-axis</p> <p>$\therefore$ slope of $AQ=\tan \theta$</p> <p>$AP$ makes angle $(\pi-\theta)$ with $x$-axis</p> <p>$\therefore \text { slope of } A P=\tan (\pi-\theta)=-\tan \theta$</p> </div> <div style='float: right; width:35%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/2e71561c-693d-424a-8b3d-82deac216400/public"></p> <!----> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\therefore$ Slope of $A Q=-$ slope of $A P$</p> <p>$ \Rightarrow \quad \frac{3-0}{5-a}=-\frac{2-0}{1-a} $</p> <p>$\Rightarrow 3(1-a)=-2(5-a) \Rightarrow \ 3-3 a=-10+2 a$</p> <p>$ \Rightarrow \quad-5 a=-13 $</p> <p>$ \Rightarrow \quad 5 a=13 \Rightarrow \ a=13 / 5 $</p> <p>$ \therefore \quad \text { Required Point }A(a, 0) $</p> <p>$ \text {i.e. }\ \ A(13 / 5,0)$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/350-2833.2.jpg"></p> </div>
### <Success>Problem 16</Success> <div style='float: left; width:70%;font-size:30px;'> <p>Find the equation of the stright line Which cuts off intercept on $x$-axis which is twice that on $y$-axis and is at a unit distance from the origin.</p> <p><strong>Solution:</strong></p> <p>Let the equation of line be</p> <p>$ \frac{x}{2 a}+\frac{y}{a}=1 $</p> <p>$\Rightarrow x+2 y=2 a$</p> <p>$\Rightarrow x+2 y-2 a=0\cdots (i)$</p> <p>According to the question, distance of line (i) from origin is 1 unit.</p> </div> <div style='float: right; width:30%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/2ff515c7-411c-4313-a535-d179119db700/public"></p> <!----> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\therefore |\frac{0+2 \times 0-2 a}{\sqrt{1^2+2^2}}|=1$</p> <p>$\Rightarrow|\frac{-2 a}{\sqrt{5}}|=1$</p> <p>$\Rightarrow\frac{2 a}{\sqrt{5}}= \pm 1$</p> <p>$\Rightarrow 2 a= \pm \sqrt{5}$</p> <p>$\Rightarrow a= \pm \frac{\sqrt{5}}{2}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/379-3012.0.jpg"></p> </div>
### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\therefore \quad$ Required equation of line are</p> <p>$ x+2y=2 a $</p> <p>$ \Rightarrow x+2 y= \pm 2 \times \frac{\sqrt{5}}{2}$</p> <p>$ \Rightarrow \quad x+2 y \pm \sqrt{5}=0$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-6-straight-lines-l-6_6-bbbmq1c8mfc/img/384-3057.6.jpg"></p> </div>
### <Success>Thank you</Success> <div> </div>