Chapter 10: Straight Lines

Ans: 

Given: Line passes through the point  $ (1, - 2) $ .

Let the intercept made by the line be  $ a $ on $ x -  $ axis and the intercept on $ y -  $ axis be  $ b $ .

So, equation of line is  $ \frac{x}{a} + \frac{y}{b} = 1 $ .

Given that  $ a = b $ .

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

It passes through  $ (1, - 2) $ , so  $ \frac{1}{a} - \frac{2}{a} = 1 $ 

 $ \Rightarrow  - \frac{1}{a} = 1 $ 

 $ \Rightarrow a =  - 1 $ 

So, equation of line will be  $ \frac{x}{{ - 1}} + \frac{y}{{ - 1}} = 1 $ 

 $ x + y =  - 1 $ 

Therefore, the equation is  $ x + y + 1 = 0 $ .

Ans: 

Given: $ (2,3) $ and  $ (3, - 1) $ .

Slope of line joining  $ (2,3) $ and  $ (3, - 1) $ is   $ \frac{{ - 1 - 3}}{{3 - 2}} =  - 4 $ .

Slope of the line perpendicular to this line is  $ \frac{{ - 1}}{{ - 4}} = \frac{1}{4} $ . 

The equation of line that passes through  $ (5,2) $ is  $ y - 2 = \frac{1}{4}\left( {x - 5} \right) $ .

$ \Rightarrow 4y - 8 = x - 5 $ 

$ \Rightarrow x - 4y + 3 = 0 $ 

Therefore, the equation is  $ x - 4y + 3 = 0 $ .

Ans: 

Given: $ y = \left( {2 - \sqrt 3 } \right)\left( {x + 5} \right) $ and  $ y = \left( {2 + \sqrt 3 } \right)\left( {x - 7} \right) $ .

Slope of line  $ y = \left( {2 - \sqrt 3 } \right)\left( {x + 5} \right) $ is   $ 2 - \sqrt 3  $ .

Slope of the line  $ y = \left( {2 + \sqrt 3 } \right)\left( {x - 7} \right) $ is  $ 2 + \sqrt 3  $ . 

 $ \theta  $ be the angle between two lines.

Using  $ \tan \theta  = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right| $ ,

 $ \Rightarrow \tan \theta  = \left| {\frac{{2 - \sqrt 3  - 2 - \sqrt 3 }}{{1 + \left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}}} \right| $ 

 $ \Rightarrow \tan \theta  = \left| {\frac{{ - 2\sqrt 3 }}{{1 + 4 - 3}}} \right| $ 

 $ \Rightarrow \tan \theta  = \left| { - \sqrt 3 } \right| $ 

 $ \Rightarrow \tan \theta  = \sqrt 3 {\text{ or }} - \sqrt 3  $ 

 $ \Rightarrow \tan \theta  = \tan {60^ \circ }{\text{ or }}\tan {120^ \circ } $ 

Therefore,  $ \theta  = {60^ \circ }{\text{ or }}\theta  = {120^ \circ } $ 

Ans: 

Given: Line passes through the point  $ (3,4) $ .

Let the intercept made by the line be  $ a $ on $ x -  $ axis and the intercept on $ y -  $ axis be  $ b $ .

So, equation of line is  $ \frac{x}{a} + \frac{y}{b} = 1 $ .

Given that  $ a + b = 14 $ .

 $ \Rightarrow b = 14 - a $ 

So,  $ \frac{x}{a} + \frac{y}{{14 - a}} = 1 $ 

This equation passes through $ (3,4) $ .

 $ \Rightarrow \frac{3}{a} + \frac{4}{{14 - a}} = 1 $ 

 $ \Rightarrow \frac{{3\left( {14 - a} \right) + 4a}}{{a\left( {14 - a} \right)}} = 1 $ 

 $ \Rightarrow 42 + a = 14a - {a^2} $ 

 $ \Rightarrow {a^2} - 13a + 42 = 0 $ 

 $ \Rightarrow \left( {a - 6} \right)\left( {a - 7} \right) = 0 $ 

 $ \Rightarrow a = 6,7 $ 

\[ \Rightarrow b = 14 - 6,14 - 7\]

\[ \Rightarrow b = 8,7\]

So, the equation of line when \[a = 6\] and \[b = 8\] is  $ \frac{x}{6} + \frac{y}{8} = 1 $  i.e. \[4x + 3y = 24\].

The equation of line when \[a = 7\] and \[b = 7\] is   $ \frac{x}{7} + \frac{y}{7} = 1 $  i.e. \[x + y = 7\].

Therefore, the equations are  $ x + y = 7 $ and \[4x + 3y = 24\].

Ans: 

Given: $ x + y = 4 $ and  $ 4x + 3y = 10 $ .

Let the point  $ ({x_1},{y_1}) $ be the point lying on  $ x + y = 4 $  i.e.  $ {x_1} + {y_1} = 4{\text{        (1)}} $ 

The distance of  $ ({x_1},{y_1}) $ from  $ 4x + 3y = 10 $ : 

 $ \Rightarrow \left| {\frac{{4{x_1} + 3{y_1} - 10}}{{\sqrt {{4^2} + {3^2}} }}} \right| = 1 $ 

 $ \Rightarrow \left| {\frac{{4{x_1} + 3{y_1} - 10}}{5}} \right| = 1 $ 

\[ \Rightarrow 4{x_1} + 3{y_1} - 10 =  \pm 5\]

Take the  $  +  $  sign,

\[ \Rightarrow 4{x_1} + 3{y_1} - 10 = 5\]

\[ \Rightarrow 4{x_1} + 3{y_1} = 15{\text{           }}...{\text{(2)}}\]

From \[{\text{(1)}}\], 

\[ \Rightarrow {y_1} = 4 - {x_1}\]

Put this value in\[{\text{(2)}}\], 

\[ \Rightarrow 4{x_1} + 3\left( {4 - {x_1}} \right) = 15\]

\[ \Rightarrow {x_1} = 3\]and \[{y_1} = 4 - 3 = 1\].

So, the point is  $ (3,1) $ .

Take the  $  -  $  sign,

\[ \Rightarrow 4{x_1} + 3{y_1} - 10 =  - 5\]

\[ \Rightarrow 4{x_1} + 3{y_1} = 5{\text{           }}...{\text{(3)}}\]

From \[{\text{(1)}}\], 

\[ \Rightarrow {y_1} = 4 - {x_1}\]

Put this value in\[{\text{(3)}}\], 

\[ \Rightarrow 4{x_1} + 3\left( {4 - {x_1}} \right) = 5\]

\[ \Rightarrow {x_1} =  - 7\]and \[{y_1} = 4 - ( - 7) = 11\].

So, the point is  $ ( - 7,11) $ .

Therefore, the points are  $ (3,1) $ and  $ ( - 7,11) $ .

Ans: 

Given: The tangent of an angle between  $ \frac{x}{a} + \frac{y}{b} = 1 $  and  $ \frac{x}{a} - \frac{y}{b} = 1 $ is  $ \frac{{2ab}}{{{a^2} - {b^2}}} $ .

Given that  $ \frac{x}{a} + \frac{y}{b} = 1{\text{     }}...{\text{(1)}} $ 

 $ \frac{x}{a} - \frac{y}{b} = 1{\text{     }}...{\text{(2)}} $ 

Slope of the line  $ \frac{x}{a} + \frac{y}{b} = 1 $ is  $  - \frac{b}{a} $ .

Slope of the line  $ \frac{x}{a} - \frac{y}{b} = 1 $ is  $ \frac{b}{a} $ .

Using  $ \tan \theta  = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right| $ ,

 $ \Rightarrow \tan \theta  = \left| {\frac{{ - \frac{b}{a} - \frac{b}{a}}}{{1 + \left( { - \frac{b}{a}} \right)\left( {\frac{b}{a}} \right)}}} \right| $ 

 $ \Rightarrow \tan \theta  = \left| {\frac{{ - \frac{{2b}}{a}}}{{1 - {{\left( {\frac{b}{a}} \right)}^2}}}} \right| $ 

 $ \Rightarrow \tan \theta  = \left| {\frac{{ - 2ab}}{{{a^2} - {b^2}}}} \right| $ 

 $ \Rightarrow \tan \theta  = \frac{{2ab}}{{{a^2} - {b^2}}} $ 

Hence verified.

Ans: 

Given: lines passing through  $ (1,2) $ .

The line makes  $ {30^ \circ } $  with  $ y -  $ axis.

So, the angle made with  $ x -  $ axis is  $ {60^ \circ } $ .

Slope of line $  = \tan {60^ \circ } = \sqrt 3  $ .

The equation of line that passes through  $ (1,2) $  and slope  $ \sqrt 3  $  is:

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

 $ \Rightarrow y - 2 = \sqrt 3 (x - 1) $ 

 $ \Rightarrow y - \sqrt 3 x + \sqrt 3  - 2 = 0 $ 

Therefore, the equation is  $ y - \sqrt 3 x + \sqrt 3  - 2 = 0 $ .

Ans: 

Given: Line passes through the point of intersection of  $ 2x + y = 5 $ and  $ x + 3y + 8 = 0 $  and parallel to  $ 3x + 4y = 7 $ .

The equation of line that passes through the point of intersection of  $ 2x + y = 5 $ and  $ x + 3y + 8 = 0 $ is  $ \left( {2x + y - 5} \right) + \lambda \left( {x + 3y + 8} \right) = 0 $ .

 $ \Rightarrow \left( {2 + \lambda } \right)x + \left( {1 + 3y} \right)\lambda  - 5 + 8\lambda  = 0 $ 

Slope of this line is  $ \frac{{ - \left( {2 + \lambda } \right)}}{{1 + 3\lambda }} $ .

Slope of  $ 3x + 4y = 7 $ is  $  - \frac{3}{4} $ .

As the two lines are parallel,  $ \frac{{ - \left( {2 + \lambda } \right)}}{{1 + 3\lambda }} =  - \frac{3}{4} $ 

 $ \Rightarrow \frac{{2 + \lambda }}{{1 + 3\lambda }} = \frac{3}{4} $ 

 $ \Rightarrow 8 + 4\lambda  = 3 + 9\lambda  $ 

 $ \Rightarrow 5\lambda  = 5 $ 

 $ \Rightarrow \lambda  = 1 $ 

The equation  $ \left( {2x + y - 5} \right) + \lambda \left( {x + 3y + 8} \right) = 0 $ becomes  $ 3x + 4y + 3 = 0 $ .

Therefore, the equation is  $ 3x + 4y + 3 = 0 $ .

Ans: 

Given:\[ax + by + 8 = 0\]and \[2x--3y + 6 = 0\].

We can write \[ax + by + 8 = 0\] as:

\[ \Rightarrow ax + by =  - 8\]

\[ \Rightarrow \frac{a}{{ - 8}}x + \frac{b}{{ - 8}}y = 1\]

\[ \Rightarrow \frac{x}{{\frac{{ - 8}}{a}}} + \frac{y}{{\frac{{ - 8}}{b}}} = 1\]

The intercepts are \[\frac{{ - 8}}{a},{\text{ }}\frac{{ - 8}}{b}\].

We can write \[2x--3y + 6 = 0\] as:

\[ \Rightarrow 2x--3y =  - 6\]

\[ \Rightarrow \frac{{2x}}{{ - 6}} - \frac{{3y}}{{ - 6}} = 1\]

\[ \Rightarrow \frac{x}{{ - 3}} + \frac{y}{2} = 1\]

The intercepts are \[ - 3,{\text{ }}2\].

From the question, \[\frac{{ - 8}}{a} = 3,{\text{ }}\frac{{ - 8}}{b} =  - 2\] i.e. \[a = \frac{{ - 8}}{3},{\text{ }}b = 4\]

Therefore,  $ a =  - \frac{8}{3} $ and  $ b = 4 $ .

Ans: 

Given:\[\left( {--5,{\text{ }}4} \right)\]

Assume \[a\]and \[b\]are intercepts on given line.

Then, the coordinates are \[A(a,0)\]and \[B(0,b)\].

Using \[x = \frac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}}\]and \[y = \frac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}\],

\[ \Rightarrow  - 5 = \frac{{1 \times 0 + 2 \times a}}{{1 + 2}}\]

\[ \Rightarrow 2a =  - 15\]

\[ \Rightarrow a = \frac{{ - 15}}{2}\]

So, \[A = \left( {\frac{{ - 15}}{2},0} \right)\].

 And \[4 = \frac{{1 \times b + 0 \times 2}}{{1 + 2}}\]

\[ \Rightarrow 4 = \frac{b}{3}\]

\[ \Rightarrow b = 12\]

So, \[B = \left( {0,12} \right)\].

The equation of line \[AB\]is \[y - {y_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)\].

\[ \Rightarrow y - 0 = \left( {\frac{{12 - 0}}{{0 + \frac{{15}}{2}}}} \right)\left( {x + \frac{{15}}{2}} \right)\]

\[ \Rightarrow y = \frac{8}{5}\left( {x + \frac{{15}}{2}} \right)\]

\[ \Rightarrow 5y = 8x + 60\]

Therefore,\[8x - 5y + 60 = 0\].