### <Success>Straight line lecture 5</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-5-straight-lines-l-5_6-kbmuxto3na0/img/014-0027.6.jpg"></p> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/20de9c0a-17a6-4c5d-61be-1a1fc2574000/public"></p> </div>

### <Success>Problem 1</Success> <div style='float: left; width:60%;font-size:30px;'> <p>Find the equation of the line which passes through the point (-4,3) and the portion of the line intercepted between the axes is divided internally in the ratio 5:3 by this point.</p> <p><strong>Solution:</strong></p> <p>$-4 = \frac{3 \times a + 5 \times 0}{5+3} = \frac{3a}{8}$</p> <p>$\Rightarrow 3a = -32$</p> <p>$\Rightarrow a = \frac{-32}{3}$</p> </div> <div style='float: right; width:40%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/232c7c99-c8c6-4da0-7a85-1e95c093d500/public"></p> <!----> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$3 = \frac{5 \times b + 3 \times 0}{5+3}$</p> <p>$\Rightarrow 5b = 24$</p> <p>$\Rightarrow b = \frac{24}{5}$</p> <p>A($\frac{-32}{3}$ ,0) & B(0, $\frac{24}{5}$)</p> <p>$\therefore$ equation of line AB</p> <p>$\frac{x}{\frac{-32}{3}} + \frac{y}{\frac{24}{5}} = 1$</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-5-straight-lines-l-5_6-kbmuxto3na0/img/043-0248.4.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ \Rightarrow \frac{3 x}{-32}+\frac{5 y}{2 y}=1 $</p> <p>$ \Rightarrow \frac{3 x}{4}-\frac{5 y}{3}=-8 $</p> <p>$ \Rightarrow \frac{9 x-20 y}{12}=-8 $</p> <p>$ \Rightarrow 9 x-20 y=-96 $</p> <p>$ \Rightarrow 9 x-20 y+96=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-5-straight-lines-l-5_6-kbmuxto3na0/img/000001.png"></p> </div>

### <Success>Problem 2</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Find the values of k for which the line</p> <p>$(k-3)x - (4-k^2)y + k^2 - 7k + 6 = 0$ is</p> <p>(i). Parrallel to x-axis (ii). Parrallel to y-axis (iii). Passes through origin.</p> <p><strong>Solution:</strong></p> <p>(i) Parallel to x-axis</p> <p>$\Rightarrow$ co-efficient of x=0</p> <p>$\Rightarrow k-3 = 0$</p> <p>$\Rightarrow k = 3$</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-5-straight-lines-l-5_6-kbmuxto3na0/img/000002.png"></p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>(ii) Parallel to y-axis</p> <p>$\Rightarrow$ co-efficient of y = 0</p> <p>$\Rightarrow 4 - k^2 = 0$</p> <p>$\Rightarrow k^2 = 4$</p> <p>$\Rightarrow k = \pm 2$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>(iii) Passes through origin y = mx</p> <p>y = mx + c $\Rightarrow$ c = 0</p> <p>c = 0</p> <p>$\Rightarrow k^2 - 7k + 6 = 0$</p> <p>$\Rightarrow k^2 - 6k -k + 6 = 0$</p> <p>$\Rightarrow k(K-6) -1(k-6) = 0$</p> <p>$\Rightarrow (k-1)(k-6) = 0$</p> <p>$\Rightarrow k = 1 or 6 $</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/098bf46f-6c6e-43c4-8928-ecf9c06b6500/public"></p> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/94d1c3e3-ad5b-428e-0d76-ac9ef1c18c00/public"></p> </div>

### <Success>Problem 3</Success> <div style='float: left; width:70%;font-size:30px;'> <p>Find the equation of one of the sides of an isosceler right angled traingle whose hypotenuse is given by 3x+4y=4 and the opposite vertex of the hypotenuse is (2,2).</p> <p><strong>Solution:</strong></p> <p>Slope of AB = $- \frac{3}{4}$</p> <p>Let Slope of AC = m</p> <p>$\tan 45^0 = |\frac{m-m_1}{1+mm_1}|$</p> <p>$\Rightarrow 1 = |\frac{m-(\frac{-3}{4})} {1 + m(\frac{-3}{4})}|$</p> </div> <div style='float: right; width:30%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/175214b3-e42b-4968-63c9-1e9cb12deb00/public"></p> <!----> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\Rightarrow 1 = |\frac{\frac{4m+3}{4}}{\frac{4-3m}{4}}|$</p> <p>$\Rightarrow 1 = |\frac{4m+3}{4-3m}|$</p> <p>$\Rightarrow \frac{4m+3}{4-3m} = \pm 1$</p> <p>$\Rightarrow 4 m + 3 = 4-3m$ or $4 m + 3 = - 4 + 3 m $</p> <p>$\Rightarrow 7m = 1$ or $m = -7$</p> <p>$\Rightarrow m - \frac{1}{7}$ or $m = -7$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/19475e14-da06-495f-ca0c-741b94affd00/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Equation of line with slope $m=\frac{1}{7}$</p> <p>$ y-2=\frac{1}{7}(x-2) $</p> <p>$\Rightarrow 7 y-14=x-2$</p> <p>$\Rightarrow x-7 y+12=0$</p> <p>Again, Equation of line with slope $m=-7$</p> <p>$y-2=-7(x-2) $</p> <p>$\Rightarrow y-2=-7 x + 14 $</p> <p>$\Rightarrow 7x + y - 16=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-5-straight-lines-l-5_6-kbmuxto3na0/img/000003.png"></p> </div>

### <Success>Problem 4</Success> <div style='float: left; width:60%;font-size:30px;'> <p>If one diagonal of a square is along the line $8x-15y=0$ and one of its vertex is at (1,2), then find the equations of sides of the square passing this vertex.</p> <p><strong>Solution:</strong></p> <p>Slope of diagonal = $\frac{8}{15} (m_1)$</p> <p>Let the slope of side AB = m</p> <p>$\tan 45^0 = |\frac{m-m_1}{1+mm_1}|$</p> <p>$\Rightarrow 1 =| \frac{m- \frac{8}{15}}{1 + m \times \frac{8}{15}}|$</p> </div> <div style='float: right; width:40%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/4801e0cd-fd61-4f08-8cd1-70fa1d3d9c00/public"></p> <!----> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ \Rightarrow 1=|\frac{15 m-8}{15+8 m}| $</p> <p>$ \Rightarrow \frac{15 m-8}{15+8 m}= \pm 1 $</p> <p>$ \Rightarrow 15 m-8=15+8 m $</p> <p>$ 15 m-8=-15-8 m $</p> <p>$ \Rightarrow 7 m=23 \text { or } 23 m=-7$</p> <p>$ \Rightarrow m=\frac{23}{7} \text { or } m= \frac{-7}{23}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/b71294f0-b380-4df9-4453-8fca9c2f6200/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Equation of side of square with slope</p> <p>$m=\frac{23}{7}$ and passing through A(1,2)</p> <p>$ y-2=\frac{23}{7}(x-1) $</p> <p>$ \Rightarrow 7 y-14=23 x-23 $</p> <p>$ \Rightarrow 23 x-7 y-9=0 $</p> <p>Again,equation of side with slope $m=-\frac{7}{23}$</p> <p>and passing thin $A(1,2)$ is</p> <p>$y-2=-\frac{7}{23}(x-1)$</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-5-straight-lines-l-5_6-kbmuxto3na0/img/000005.png"></p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <!--4. If one diagonal of a square is along the line $8x-15y=0$ and one of its vertex is at (1,2), then find the equations of sides of the square passing this vertex.--> <!--Slope of diagonal = $\frac{8}{15} (m_1)$--> <!--Let the slope of side AB = m--> <!--$\tan 45^0 = |\frac{m-m_1}{1+mm_1}|$--> <!--$\Rightarrow 1 =| \frac{m- \frac{8}{15}}{1 + m \times \frac{8}{15}}|$--> <p>$\Rightarrow 23 y - 46 = -7 x + 7$</p> <p>$\Rightarrow 7 x + 23 y - 53 = 0$</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-5-straight-lines-l-5_6-kbmuxto3na0/img/171-1380.0.jpg"> <img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-5-straight-lines-l-5_6-kbmuxto3na0/img/173-1388.4.jpg"></p> </div>

### <Success>Problem 5</Success> <div style='float: left; width:70%;font-size:30px;'> <p>Find the image of the point (3,8) with respect to the $x+3y=7$, assuming the line to be a plane mirror.</p> <p><strong>Solution:</strong></p> <p>Slope of l is $-\frac{1}{3}$</p> <p>$\therefore$ slope of PQ $\perp$ l</p> <p>=$ - \frac{1}{\frac{-1}{3}} = 3$</p> <p>Equation of line PQ with slope 3 and passing through P(3,8)</p> <p>$y-8=3(x-3) \ \Rightarrow 3x-y-1=0$</p> </div> <div style='float: right; width:30%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/fb4358d4-0864-4b12-e492-4c15f59b9e00/public"></p> <!----> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Given equation of line $x+3 y=7$</p> <p>$\Rightarrow x=7-3 y$</p> <p>put $x=7-3 y$ in (i)</p> <p>$ 3(7-3 y)-y-1=0 $</p> <p>$\Rightarrow 21-9 y-y-1=0 \Rightarrow -10 y+20=0 $</p> <p>$\Rightarrow y=2 $</p> <p>$\therefore x=7-3 y=7-3 \times 2=7-6=1$</p> <p>$\therefore$ Co-ordinate of foot of $\perp$ N is (1,2)</p> <!--5. Find the image of the point (3,8) with respect to the $x+3y=7$, assuming the line to be a plane mirror.--> <!--Slope of l is $-\frac{1}{3}$--> <!--$\therefore$ slope of PQ $\perp$ l--> <!-- =$ - \frac{1}{\frac{-1}{3}} = 3$--> <!--Equation of line PQ with slope 3 and passing through P(3,8)--> <!--$y-8=3(x-3)$--> <!--$\Rightarrow 3x-y-1=0$--> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/255019e4-fdf1-436b-de0a-6c6523624b00/public"></p> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-11-chapter-5-straight-lines-l-5_6-kbmuxto3na0/img/198-1742.4.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <!--5. Find the image of the point (3,8) with respect to the $x+3y=7$, assuming the line to be a plane mirror.--> <!--Slope of l is $-\frac{1}{3}$--> <!--$\therefore$ slope of PQ $\perp$ l--> <!-- =$ - \frac{1}{\frac{-1}{3}} = 3$--> <!--Equation of line PQ with slope 3 and passing through P(3,8)--> <!--$y-8=3(x-3)$--> <!--$\Rightarrow 3x-y-1=0$--> <p>Since, $N$ is the mid point of PQ</p> <p>$\therefore \frac{\alpha+3}{2} =1 \Rightarrow \alpha+3=2 \Rightarrow \alpha=-1$</p> <p>$\frac{\beta+8}{2} =2 \Rightarrow \beta+8=4 \Rightarrow \beta=-4 $</p> <p>$\therefore$ Image of point P(3,8) with respect to a line</p> <p>$x+3y=7$ is Q(-1,-4)</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-5-straight-lines-l-5_6-kbmuxto3na0/img/212-1891.2.jpg"></p> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/16b5cab3-e44c-498e-2d0b-05ff4b8ec000/public"></p> </div>

### <Success>Problem 6</Success> <div style='float: left; width:70%;font-size:26px;'> <p>If the area of the traingle formed by a line with co-ordinate axes is $54\sqrt{3}$ square unit and the perpendicular drawn from the origin to the line makes an angle of $60^0$ with x-axis, find the equation of the line.</p> <p><strong>Solution:</strong></p> <p>Given area ($\Delta OAB$) = $54\sqrt{3}$</p> <p>Let the equation of line be</p> <p>$\frac{x}{a} + \frac{y}{b} = 1$</p> <p>Equation of line in normal for</p> <p>$x \cos \alpha + y \sin \alpha = p$</p> <p>$\Rightarrow x \cos 60^0 + y \sin 60^0 = p$</p> </div> <div style='float: right; width:30%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/37cd554d-6121-4f2a-ca41-17330e98b600/public"></p> <!----> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:26px;'> <p>$ \Rightarrow x \times \frac{1}{2}+y \times \frac{\sqrt{3}}{2}=p $</p> <p>$ \Rightarrow \frac{x}{2 p}+\frac{y}{\frac{2 p}{\sqrt{3}}}=1$–(ii)</p> <p>on Comparring (i) & (ii), we will have</p> <p>$a=2 p, b=\frac{2 p}{\sqrt{3}}$</p> <p>Given, area $(\triangle O A B)=54 \sqrt{3}$ sq.unit</p> <p>$ \Rightarrow \frac{1}{2} \times a \times b=54 \sqrt{3} $</p> <p>$ \Rightarrow \frac{1}{2} \times 2 p \times \frac{2 b}{\sqrt{3}}=54 \sqrt{3}$</p> <p>$ \Rightarrow 2p^2 = 54 \times 3$</p> <p>$ \Rightarrow p^2 = 81$</p> <p>$p = \pm 9$</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-5-straight-lines-l-5_6-kbmuxto3na0/img/000006.png"></p> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/bed413ab-8de6-4a35-3216-fdbf9eb10000/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Equation of line with $\alpha$ = $60^0$ & $p= \pm3$</p> <p>$ x \cos \alpha+y \sin \alpha=p $</p> <p>$\Rightarrow x \cos 60^0+y \sin 60^0=3 $</p> <p>or</p> <p>$ x \cos 60^0+y \sin 60^0=-3 \Rightarrow x \times \frac{1}{2}+y \times \frac{\sqrt{3}}{2}=3 $</p> <p>$\Rightarrow x \times \frac{1}{2}+y \times \frac{\sqrt{3}}{2}=-3 \Rightarrow x+\sqrt{3} y-6=0 $</p> <p>or</p> <p>$ x+\sqrt{3} y+6=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-5-straight-lines-l-5_6-kbmuxto3na0/img/000007.png"></p> </div>

### <Success>Problem 7</Success> <div style='float: left; width:70%;font-size:30px;'> <p>Find the distance of the point P(4,1) from the line $4x-y=0$ measured along the line making an angle of $135^0$ with the positive direction of x-axis.</p> <p><strong>Solution:</strong></p> <p>Given equation of line l.</p> <p>$4x-y=0$ (eqn-1)</p> <p>Also given line $l_1$ makes angle $135^0$ with x-axis</p> <p>$\therefore$ slope of $l_1 = \tan 135^0 = -1$</p> </div> <div style='float: right; width:30%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/0ad009e2-8668-4db3-92da-8d65649bf300/public"></p> <!---->

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$y-1=-1(x-4) $</p> <p>$\Rightarrow y-1=-x+4 \Rightarrow x+y-5=0 $ (eqn-2)</p> <p>$4 x-y=0 $ (eqn-1)</p> <p>$y=4 x$</p> <p>put $y=4 x$ in (eqn-2)</p> <p>$x+4 x-5=0 $</p> <p>$\Rightarrow 5x=5 \Rightarrow x=1 $</p> <p>$\therefore y=4$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/acd906c1-11d3-44d9-bb0a-5947002c7600/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Distance of line l from P(4,1) along the line $l_1$ is PQ</p> <p>$\therefore PQ = \sqrt{(4-1)^2+ (1-4)^2}$</p> <p>= $\sqrt{3^2 + 3^2}$</p> <p>= $3\sqrt{2}$ unit.</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-5-straight-lines-l-5_6-kbmuxto3na0/img/305-2697.6.jpg"></p> </div>

### <Success>Problem 8</Success> <div style='float: left; width:70%;font-size:30px;'> <p>8.If the line 3x+y-2=0, px + 2y - 3 = 0 and 2x-y-3 = 0 are concurrent, find the value of p</p> </div>

### <Success>Solution</Success> <div style='float: left; width:60%;font-size:30px;'> <p>If $a_1 x + b_1 y +c_1=0$</p> <p>$a_2 x + b_2 y + c_2 = 0$</p> <p>$a_3 x + b_3 y + c_3 = 0$</p> <p>lines are concurrent, then</p> <p>$\left|\begin{array}{ccc}a_1 & b_1 & c_1 \\a_2 & b_2 & c_3 \\a_3 & b_3 & c_3\end{array}\right|$ $=0$</p> </div> <div style='float: right; width:40%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/e9d98171-3c90-416f-4415-d9a8bda47400/public"></p> <!----> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ a_1(b_2 c_3-b_3 c_2)-b_1(a_2 c_3-a_3 c_2) + c_1(a_1 b_3-a_3 b_2)=0$</p> <p>Given equation of lines are</p> <p>$ 3 x+y-2=0$ (i)</p> <p>$ p x+2 y-3=0$ (i)</p> <p>$ 2 x-y-3=0$ (iii)</p> <p>Since Given line (i), (ii) & (iii) are concurrent</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-5-straight-lines-l-5_6-kbmuxto3na0/img/000008.png"></p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\therefore$ $\left|\begin{array}{ccc}3 & 1 & -2 \\p & 2 & -3 \\2 & -1 & -3\end{array}\right|$ $=0$</p> <p>$ \Rightarrow 3(-6-3)-1(-3 p+6)-2(-p-4)=0 $</p> <p>$ \Rightarrow-27+3 p-6+2 p+8=0 $</p> <p>$ \Rightarrow 5 p-33+8=0 $</p> <p>$ \Rightarrow 5 p-25=0 $</p> <p>$ \Rightarrow p=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-5-straight-lines-l-5_6-kbmuxto3na0/img/366-3109.2.jpg"></p> </div>

### <Success>Thank you</Success> <div> </div>