Chapter 10: Straight Lines

Ans: 

Given: Line makes  $ {120^ \circ } $ with the positive direction of  $ x -  $ axis. 

The line makes  $ {120^ \circ } $  with the positive direction of  $ x -  $ axis.

So, slope of the line is  $ \tan {120^ \circ } =  - \sqrt 3  $ .

Equation of the required line is: 

 $ y =  - \sqrt 3 x + c $ 

 $ \Rightarrow \sqrt 3 x + y - c = 0 $ . 

This line is 4 units away from  $ (0,0) $ .

 $ \Rightarrow \frac{{\left| {\sqrt 3 (0) + 0 - c} \right|}}{{\sqrt {3 + 1} }} = 4 $ 

 $ \Rightarrow \quad |c| = 8 $ 

 $ \Rightarrow c =  \pm 8 $ 

Therefore, the equation of line is  $ \sqrt 3 x + y \pm 8 = 0 $ .

Ans: 

Given: $ 3x + 4y + 4 = 0 $ 

The hypotenuse is along the line  $ 3x + 4y + 4 = 0 $ 

Slope of the hypotenuse is  $ \frac{{ - 3}}{4} $ .

As the triangle is isosceles right angled triangle, two angles will be equal to  $ {45^\circ } $ .

Let the slope of the line making  $ {45^\circ } $  with the hypotenuse be  $ m $ .

 $ \Rightarrow \tan {45^\circ } = \left| {\frac{{m - \left( { - \frac{3}{4}} \right)}}{{1 + m\left( { - \frac{3}{4}} \right)}}} \right| $ 

 $ \Rightarrow \tan {45^\circ } = \left| {\frac{{4m + 3}}{{4 - 3m}}} \right| $ 

 $ \Rightarrow \frac{{4m + 3}}{{4 - 3m}} =  \pm 1 $ 

 $ \Rightarrow 4m + 3 = 4 - 3m $ or $ 4m + 3 = 3m - 4 $ 

 $ \Rightarrow m = \frac{1}{7} $ or $ m =  - 7 $ 

Slope of this line is  $ \frac{1}{7} $ , so the equation of this line is  $ y - 2 = (1/7)(x - 2) $  i.e.  $ x - 7y + 12 = 0 $ .

The slope of line perpendicular to it is  $  - 7 $ . 

The equation of this line is:  $ y - 2 =  - 7(x - 2) $  i.e.  $ 7x + y - 16 = 0 $ .

Therefore, the equation of required line is  $ 7x + y - 16 = 0 $ .

Ans: 

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

The altitude from the vertex  $ (2, - 1) $ on the base  $ x + y - 2 $  meets at the midpoint of base. 

The length of altitude is the distance of  $ (2, - 1) $  from $ x + y - 2 $  $  = \left| {\frac{{2 + ( - 1) - 2}}{{\sqrt {{1^2} + {1^2}} }}} \right| = \frac{1}{{\sqrt 2 }} $ .

The triangle is an equilateral triangle. So,  $ \sin {60^\circ } = \frac{{{\text{Altitude}}}}{{{\text{side of triangle}}}} $ .

$ \Rightarrow {\text{length of side of triangle}} = \frac{{{\text{Altitude}}}}{{\sin {{60}^\circ }}} = \frac{{\frac{1}{{\sqrt 2 }}}}{{\frac{{\sqrt 3 }}{2}}} = \sqrt {\frac{2}{3}}  $ .

Therefore, the length of side of the triangle is  $ \sqrt {\frac{2}{3}}  $ .

Ans: 

Given:  $ (2,0),(0,2) $  and  $ (1,1) $ 

Assume that the variable line passing through the fixed point  $ P $ is  $ ax + by + c = 0 $ .

Perpendicular distance from the point $ A(2,0) = \frac{{2a + 0 + c}}{{\sqrt {{a^2} + {b^2}} }} $ 

Perpendicular distance from the point $ B(0,2) = \frac{{0 + 2b + c}}{{\sqrt {{a^2} + {b^2}} }} $ 

Perpendicular distance from the point $ C(1,1) = \frac{{a + b + c}}{{\sqrt {{a^2} + {b^2}} }} $ 

Given that,  $ \frac{{2a + 0 + c}}{{\sqrt {{a^2} + {b^2}} }} + \frac{{0 + 2b + c}}{{\sqrt {{a^2} + {b^2}} }} + \frac{{a + b + c}}{{\sqrt {{a^2} + {b^2}} }} = 0 $ 

 $ \Rightarrow \quad 3a + 3b + 3c = 0 $ or $ a + b + c = 0 $ 

Therefore, the variable line passes through the fixed point  $ (1,1) $ .

Ans: 

Given:  $ x + y = 4 $ and  $ (1,2) $ .

Let the slope of line be  $ m $ . 

It passes through $ A(1,2) $ 

 So, the equation of line is  $ y - 2 = m(x - 1) $  or  $ mx - y + 2 - m = 0{\text{       }}...{\text{(1)}} $ 

The equation of the given line is  $ x + y - 4 = 0\quad ...(2) $ 

Assume that these lines meet at point  $ B $ .

Solving  $ (1) $ and  $ (2) $ , 

$ B \equiv \left( {\frac{{m + 2}}{{m + 1}},\frac{{3m + 2}}{{m + 1}}} \right) $ 

Given that  $ AB = \frac{{\sqrt 6 }}{3} $ 

$ \Rightarrow \quad A{B^2} = \frac{6}{9} $ 

$ \Rightarrow {\left( {\frac{{m + 2}}{{m + 1}} - 1} \right)^2} + {\left( {\frac{{3m + 2}}{{m + 1}} - 2} \right)^2} = \frac{6}{9} $ 

$ \Rightarrow \quad {\left( {\frac{1}{{m + 1}}} \right)^2} + {\left( {\frac{m}{{m + 1}}} \right)^2} = \frac{2}{3} $ 

$ \Rightarrow \frac{{1 + {m^2}}}{{{{(1 + m)}^2}}} = \frac{2}{3} $ 

$ \Rightarrow \quad 3 + 3{m^2} = 2 + 2{m^2} + 4m $ 

$ \Rightarrow {m^2} - 4m + 1 = 0 $ 

$ \Rightarrow m = \frac{{4 \pm \sqrt {16 - 4} }}{2} = 2 \pm \sqrt 3  = 2 + \sqrt 3  $ or $ 2 - \sqrt 3  $ 

$ \Rightarrow \tan \theta  = 2 + \sqrt 3  $ or $ 2 - \sqrt 3  $ 

$ \Rightarrow \theta  = {75^\circ } $ or  $ \theta  = {15^\circ } $ .

Therefore,  $ \theta  = {75^ \circ }{\text{ or }}\theta  = {15^ \circ } $ 

Ans: 

Given:  $ \frac{1}{a} + \frac{1}{b} =  $  constant

The equation of line in intercept form is  $ \frac{x}{a} + \frac{y}{b} = 1 $ .

As  $ \frac{1}{a} + \frac{1}{b} =  $  constant, 

Let  $ \frac{1}{a} + \frac{1}{b} $  $  = \frac{1}{k} $ . 

So,  $ \frac{k}{a} + \frac{k}{b} = 1 $ 

Thus,  $ (k,k) $  lies on  $ \frac{x}{a} + \frac{k}{b} = 1 $ .

Therefore, the line passes through the fixed point  $ (k,k) $ .

$\Rightarrow \therefore  - 4 = \frac{{5 \times 0 + 3a}}{{5 + 3}}$  

$\Rightarrow  - 4 = \frac{{3a}}{8}\left[ {\because X = \frac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}},y = \frac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}} \right]$ 

$\Rightarrow 3a =  - 32$                                                                                                              

$\Rightarrow a = \frac{{ - 32}}{3}$   and  $3 = \frac{{5 \times b + 3 \times 0}}{{5 + 3}}$                                                                

$\Rightarrow 3 = \frac{{5 \times b}}{8} \Rightarrow 5b = 24 \Rightarrow b = \frac{{25}}{5}$  

Intercept form of the line is  $\frac{x}{{\frac{{ - 32}}{3}}} + \frac{y}{{\frac{{24}}{5}}} = 1$                      

$\Rightarrow \frac{{ - 3x}}{{32}} + \frac{{5y}}{{24}} = 1$ 

$\Rightarrow  - 9x + 20y = 96 \Rightarrow 9x - 20y + 96 = 0$  

Hence, The required equation is  $9x - 20y + 96 = 0$ 

Ans: Given equation are

$\mathbf{x}-\mathbf{y}+\mathbf{1}=\mathbf{0} \ldots(\mathbf{i}) $  And  $ \mathbf{2} \mathbf{x}-\mathbf{3} \mathbf{y}+\mathbf{5}=\mathbf{0} \ldots(\mathbf{i i})$  Solving eq. (i) and eq. (ii) 

we get  $2 x-2 y+2=0$   $2 x-3 y+5=0$   

$y-3=0, y=3$ 

From eq. (i) we have 

$\mathbf{x}-\mathbf{3}+\mathbf{1}=\mathbf{0} \Rightarrow \mathbf{x}=\mathbf{2}$ 

So,  $(2,3)$  is the point of intersection of eq.(i) and eq. (ii). 

Let m be the slope of the required line

 $\therefore$  Equation of the line is

 $y-3=m(x-2)$ 

 $\Rightarrow y-3=m x-2 m$ 

 $\Rightarrow m x-y+3-2 m=0$ 

Since, the perpendicular distance from  $(3,2)$ 

to the line is  $\frac{7}{5}$  then,

 $\frac{7}{5}=\left|\frac{m(3)-2+3-2 m}{\sqrt{m^{2}+1}}\right|$ 

 $\Rightarrow \frac{49}{25}=\frac{(3 m-2+3-2 m)^{2}}{m^{2}+1}$ 

 $\Rightarrow \frac{49}{25}=\frac{(m+1)^{2}}{m^{2}+1}$ 

 $\Rightarrow 49 m^{2}+49=25 m^{2}+50 m+25$ 

 $\Rightarrow 49 m^{2}-25 m^{2}-50 m+49-25=0$ 

 $\Rightarrow 24 m^{2}-50 m+24=0$ 

 $\Rightarrow 12 m^{2}-25 m+12=0$ 

 $\Rightarrow 12 m^{2}-16 m-9 m+12=0$ 

 $\Rightarrow 4 m(3 m-4)-3(3 m-4)=0$ 

 $\Rightarrow(3 m-4)(4 m-3)=0$ 

 $\Rightarrow 3 m-4=0 \text { and } 4 m-3=0$ 

 $\Rightarrow 3 m-4=0 \text { and } 4 m-3=0$ 

 $\therefore m=\frac{4}{3}, \frac{3}{4}$ 

Equation of the line taking  $ m=\frac{4}{3}$  is

 $y-3=\frac{4}{3}(x-2)$ 

 $\Rightarrow 3 y-9=4 x-8 \Rightarrow 4 x-3 y+1=0$ 

Equation of the line taking  $m=\frac{3}{4}$  is

 $y-3=\frac{3}{4}(x-2)$ 

 $\Rightarrow 4 y-12=3 x-6 \Rightarrow 3 x-4 y+6=0$ 

Hence, the required equation are

 $4 x-3 y+1=0 \text { and } 3 x-4 y+6=0$ 

Ans: Let coordinates of a moving point P be  $ (x,y). $  

Given that the sum of the distance from the axes to the point is always 1

$ {\therefore \left| x \right|{\text{ }} + {\text{ }}\left| y \right|{\text{ }} = {\text{ }}1} $ 

$ { \Rightarrow x{\text{ }} + {\text{ }}y{\text{ }} = {\text{ }}1} $  

$ { \Rightarrow  - x--y = 1} $  

$ { \Rightarrow  - x + y = 1} $ 

$ { \Rightarrow x--y{\text{ }} = {\text{ }}1} $ 

Hence, these equations gives us the locus of the point P which is a square.

Ans: Given lines are  $ y - \sqrt 3 \left| x \right| = 2 $ 

$ \Rightarrow y - \sqrt 3  x = 2 $   if x ≥ 0 …(i)

And  $ y + \sqrt 3  x = 2 $  if x < 0 …(ii)

Slope of eq. (i) is  $ \tan \theta  = \sqrt 3  $    $ \therefore \theta  = {60^o} $  

Slope of eq. (ii) is  $ \tan \theta  =  - \sqrt 3  $    $ \therefore \theta  = {120^o} $ 

120° Solving eq. (i) and eq. (ii) we get

$ y - \sqrt 3 x = 2 $ 

$ y + \sqrt 3 x = 2 $ 

$ \_\_\_\_\_\_\_\_\_\_ $ 

$ 2y = 4 \Rightarrow y = 2 $  

Putting the value of y is eq. (i) we get

$ x = 0 $  

Point of intersection of line (i) and (ii) is  $ Q(0,2) $  

$ \therefore QO = 2 $  

$  $  In  $ \Delta PEQ, $                                                                                                 

$ \cos {30^o} = \frac{{PQ}}{{QE}} \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{{PQ}}{5} $ 

$ \Rightarrow \therefore PQ = \frac{{5\sqrt 3 }}{2} $ 

$ \therefore OP = OQ + PQ $  

$  = 2 + \frac{{5\sqrt 3 }}{2} $  

Hence, the coordinates of the foot of perpendicular  $  = (0,2 + \frac{{5\sqrt 3 }}{2}) $