Chapter 10: Straight Lines

Ans: Given equation is $\frac{x}{a}+\frac{y}{b}=1$

Since, $\mathrm{p}$ is the length of perpendicular drawn from the origin to the given line

$\therefore p=\left|\frac{\frac{0}{a}+\frac{0}{b}-1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}\right|$

Squaring both sides, we have

$p^{2}=\frac{1}{\frac{1}{a^{2}}+\frac{1}{b^{2}}}$ $\Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{p^{2}} \ldots \ldots(i)$ Since $a^{2}, p^{2}, b^{2}$ are in A.P. $\therefore 2 p^{2}=a^{2}+b^{2}$ $\Rightarrow p^{2}=\frac{a^{2}+b^{2}}{2} \Rightarrow \frac{1}{p^{2}}=\frac{2}{a^{2}+b^{2}}$

Putting the value of $\frac{1}{p^{2}}$ is eq. (i) we get,

$\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{2}{a^{2}+b^{2}}$

$\Rightarrow \frac{a^{2}+b^{2}}{a^{2} b^{2}}=\frac{2}{a^{2}+b^{2}}$

$\Rightarrow\left(a^{2}+b^{2}\right)^{2}=2 a^{2} b^{2}$

$\Rightarrow a^{4}+b^{4}+2 a^{2} b^{2}=2 a^{2} b^{2}$

$\Rightarrow a^{4}+b^{4}=0$

Hence proved.

Ans: 

Given: A line cuts off an intercept  $  - 3 $  from the  $ y -  $ axis.

Let the equation of required line be  $ y = mx + c $ .

As  $ c =  - 3 $ and  $ m = \frac{3}{5} $ ,

The equation of line is:  $ y = \frac{3}{5}x - 3 $ 

 $ \Rightarrow 5y - 3x + 15 = 0 $ 

Hence, the equation is  $ 5y - 3x + 15 = 0 $ .

Correct Option : A

Ans: 

Given: A line cuts off intercepts of equal lengths on the axes.

The equation of required line is $ \frac{x}{a} + \frac{y}{a} = 1 $ .

$ \Rightarrow x + y = a $ 

$ \Rightarrow y =  - x + a $ 

Slope $  =  - 1 $ 

Hence, the slope is  $  - 1 $ .

Correct Option: A

Ans: 

Given: $ (3,2) $ and  $ y = x $ .

Slope of  $ y = x $ is  $ 1 $ .

So, the slope of line perpendicular to it is  $  - 1 $ .

The line passes through  $ (3,2) $ .

So, equation of required line is  $ y - 2 =  - 1(x - 3) $  i.e.  $ x + y = 5 $ .

Hence, equation is  $ x + y = 5 $ .

Correct Option: B

Ans: 

Given: $ (1,2) $ and  $ x + y + 1 = 0 $ .

Slope of  $ x + y + 1 = 0 $ is  $  - 1 $ .

So, the slope of line perpendicular to it is  $ 1 $ .

The line passes through  $ (1,2) $ .

So, equation of required line is  $ y - 2 = 1(x - 1) $  i.e.  $ y - x - 1 = 0 $ .

Hence, equation is  $ y - x - 1 = 0 $ .

Correct Option: B

Ans: 

Given: The intercepts on the axes are  $ a,{\text{ }} - b $ and  $ b,{\text{ }} - a $ .

The intercepts of line are  $ a $ and  $  - b $ .

 So, the line passes through the points  $ (a,0) $ and  $ (0, - b) $ .

Thus, the slope of line is  $ {m_1} = \frac{{ - b - 0}}{{0 - a}} = \frac{b}{a} $ .

The intercepts of line are  $ b $ and  $  - a $ .

 So, the line passes through the points  $ (b,0) $ and  $ (0, - a) $ .

Thus, the slope of line is  $ {m_2} = \frac{{ - a - 0}}{{0 - b}} = \frac{a}{b} $ .

Let  $ \theta  $  be the angle between the lines.

 $ \Rightarrow \tan \theta  = \frac{{\frac{b}{a} - \frac{a}{b}}}{{1 + \frac{a}{b} \cdot \frac{b}{a}}} $ 

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

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

Hence,  $ \tan \theta  = \frac{{{b^2} - {a^2}}}{{2ab}} $ .

Correct Option: C

Ans: 

Given: $ (2, - 3) $ and  $ (4, - 5) $ .

The points  $ (2, - 3) $  and  $ (4, - 5) $  lie on  $ \frac{x}{a} + \frac{y}{b} = 1 $ .

 $ \Rightarrow \frac{2}{a} - \frac{3}{b} = 1{\text{         }}...{\text{(1)}} $ 

And  $ \frac{4}{a} - \frac{5}{b} = 1{\text{               }}...{\text{(2)}} $ 

Multiply  $ {\text{(1)}} $  by  $ {\text{2}} $  and subtract  $ {\text{(2)}} $ from it.

 $ \Rightarrow  - \frac{6}{b} + \frac{5}{b} = 1 $ 

 $ \Rightarrow \frac{{ - 1}}{b} = 1 $ 

 $ \Rightarrow b =  - 1 $ 

Put  $ b =  - 1 $  in $ {\text{(1)}} $ ,

 $ \Rightarrow \frac{2}{a} + 3 = 1 $ 

 $ \Rightarrow a =  - 1 $ 

So,  $ (a,b) = ( - 1, - 1) $ 

Hence,  $ (a,b) = ( - 1, - 1) $ .

Correct Option: D

Ans: 

Given: $ 2x - 3y + 5 = 0 $ and  $ 3x + 4y = 0 $ .

Solve the lines  $ 2x - 3y + 5 = 0 $ and  $ 3x + 4y = 0 $ .

The point of intersection is  $ \left( {\frac{{ - 20}}{{17}},\frac{{15}}{{17}}} \right) $ .

The distance of  $ \left( {\frac{{ - 20}}{{17}},\frac{{15}}{{17}}} \right) $ from  $ 5x - 2y = 0 $  is  $ \frac{{\left| {5 \times \left( { - \frac{{20}}{{17}}} \right) - 2\left( {\frac{{15}}{{17}}} \right)} \right|}}{{\sqrt {25 + 4} }} $ 

The distance of  $ \left( {\frac{{ - 20}}{{17}},\frac{{15}}{{17}}} \right) $ from  $ 5x - 2y = 0 $  is  $ \frac{{\left| {\frac{{ - 100}}{{17}} - \frac{{30}}{{17}}} \right|}}{{\sqrt {29} }} $ 

The distance of  $ \left( {\frac{{ - 20}}{{17}},\frac{{15}}{{17}}} \right) $ from  $ 5x - 2y = 0 $  is  $ \frac{{130}}{{17\sqrt {29} }} $ 

Hence, the distance is  $ \frac{{130}}{{17\sqrt {29} }} $ .

Correct Option: A

Ans: 

Given: $ (3, - 2) $ and  $ \sqrt 3 x + y = 1 $ .

The slope of $ \sqrt 3 x + y = 1 $ is  $ {m_1} =  - \sqrt 3  $ . 

Let the slope of required line which makes  $ {60^\circ } $  with  $ \sqrt 3 x + y = 1 $ be  $ m $ .

Then,  $ \tan {60^\circ } = \left| {\frac{{ - \sqrt 3  - m}}{{1 - \sqrt 3 m}}} \right| $ 

 $ \Rightarrow \left| {\frac{{ - \sqrt 3  - m}}{{1 - \sqrt 3 m}}} \right| = \sqrt 3  $ 

 $ \Rightarrow  - \sqrt 3  - m = \sqrt 3  - 3m $ or\[ - \sqrt 3  - m =  - \sqrt 3  + 3m\]

 $ \Rightarrow m = \sqrt 3  $ or $ m = 0 $ 

Line passes through $ (3, - 2) $ . 

So, the equation of line is:  $ y + 2 = \sqrt 3 (x - 3) $  or  $ y + 2 = 0 $ 

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

Hence, the equations are  $ y + 2 = 0,{\text{ }}\sqrt 3 x - y - 2 - 3\sqrt 3  = 0 $ .

Correct Option: A

Ans: 

Given: Line passes through  $ (1,0) $ .

Let the slope of the required line be  $ m $ .

The equation of line passing through  $ (1,0) $  is: $ y - 0 = m(x - 1) $  i.e.  $ y - mx + m = 0 $ .

The distance of this line from  $ (0,0) $ is  $ \frac{{\sqrt 3 }}{2} $ .

 $ \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{{|0 - 0 + m|}}{{\sqrt {1 + {m^2}} }} $ 

 $ \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{{|m|}}{{\sqrt {1 + {m^2}} }} $ 

 $ \Rightarrow 3 + 3{m^2} = 4{m^2} $ 

 $ \Rightarrow {m^2} = 3 $ 

 $ \Rightarrow m =  \pm \sqrt 3  $ 

Hence, equations of lines are:  $ \sqrt 3 x + y - \sqrt 3  = 0 $ , $ \sqrt 3 x - y - \sqrt 3  = 0 $ .

Correct Option: A