Chapter 10: Straight Lines

Ans: The given statement is true.

The intersection of the lines  $ x + 2y - 10 = 0 $  and  $ 2x + y + 5 = 0 $  is their solution coordinates. Solving the two equations algebraically gives,

$ x = \left( { - \frac{{20}}{3}} \right) $ and  $ y = \left( {\frac{{25}}{3}} \right) $ , so the solution coordinate is  $ \left( {\; - \frac{{20}}{3},{\text{ }}\frac{{25}}{3}} \right) $ .

Now, if the line  $ 5x + 4y = 0 $  passes through the above coordinates then it must satisfy the point. On substitution,

$ \Rightarrow 5 \times \left( { - \frac{{20}}{3}} \right) + 4 \times \left( {\frac{{25}}{3}} \right) = 0 $ 

$ \Rightarrow \left( { - \frac{{100}}{3}} \right) + \left( {\frac{{100}}{3}} \right) = 0 $ 

$ \Rightarrow 0 = 0 $  

Hence, the point satisfies the line.

Ans: The given statement is true.

Let us assume the slope of one of the remaining two sides as  $ m $ . The slope of the given line  $ x + y = 2 $ is  $  - 1 $ . If  $ {m_1} $  and  $ {m_2} $  are slopes of two line segments then the angle  $ \left( \theta  \right) $  between them is given as  $ \tan \theta  = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right| $ .

The angle between any two sides of an equilateral triangle is  $ {60^ \circ } $ ,

$ \Rightarrow \tan {60^ \circ } = \left| {\frac{{ - 1 - m}}{{1 + \left( { - 1} \right) \times m}}} \right| $ 

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

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

Removing the modulus sign,

 $ \Rightarrow  \pm {\text{ }}\sqrt 3  = \left( {\frac{{1 + m}}{{1 - m}}} \right) $ 

1. Considering the positive sign,

\[ \Rightarrow \sqrt 3  = \left( {\frac{{1 + m}}{{1 - m}}} \right)\]

\[ \Rightarrow m = \left( {\frac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}}} \right)\]

\[ \Rightarrow m = \left( {2 - \sqrt 3 } \right)\] (By rationalization)

1. Considering the negative sign,

$ \Rightarrow  - \left( {\sqrt 3 } \right) = \left( {\frac{{1 + m}}{{1 - m}}} \right) $ 

$ \Rightarrow m = \left( {\frac{{\sqrt 3  + 1}}{{\sqrt 3  - 1}}} \right) $  

\[ \Rightarrow m = \left( {2 + \sqrt 3 } \right)\] (By rationalization)

Hence, there are two values of  $ m $  for the two remaining sides of the triangle. Both the sides will pass through the point  $ \left( {2,{\text{ }}3} \right) $ . So, equation of the other two sides are  $ \left( {y - 3} \right) = \left( {2 \pm \sqrt 3 } \right)\left( {x - 2} \right) $ .

Ans: The given statement is true.

Solving the equations  $ 4x + y - 1 = 0 $  and  $ 7x - 3y - 35 = 0 $  algebraically gives  $ x = 2 $ and  $ y =  - 7 $ . So the point of intersection is  $ \left( {2,{\text{ }} - 7} \right) $ .

The equation of the line passing through the points \[\left( {3,{\text{ }}5} \right)\] and  $ \left( {2,{\text{ }} - 7} \right) $  is,

$ \Rightarrow \left( {y - 5} \right) = \left( {\frac{{ - 7 - 5}}{{2 - 3}}} \right)\left( {x - 3} \right) $ 

$ \Rightarrow \left( {y - 5} \right) = \left( {12} \right)\left( {x - 3} \right) $ 

$ \Rightarrow 12x - y - 31 = 0 $  

Now, the distance of the point  $ \left( {0,{\text{ }}0} \right) $  from the line  $ 12x - y - 31 = 0 $ ,

$ = \left| {\frac{{\left( {0 \times 12} \right) + \left( {0 \times \left( { - 1} \right)} \right) - 31}}{{\sqrt {{{12}^2} + {{\left( { - 1} \right)}^2}} }}} \right| $ 

$ = \frac{{31}}{{\sqrt {145} }} $  

Similarly, the distance of the point  $ \left( {8,{\text{ }}34} \right) $  from the line  $ 12x - y - 31 = 0 $ ,

$ = \left| {\frac{{\left( {8 \times 12} \right) + \left( {34 \times \left( { - 1} \right)} \right) - 31}}{{\sqrt {{{12}^2} + {{\left( { - 1} \right)}^2}} }}} \right| $ 

$ = \frac{{31}}{{\sqrt {145} }} $  

Hence, the points are equidistant from the required line.

Ans: The given statement is true.

Let the coordinates of the foot of perpendicular drawn from the origin $(0,0)$ is $(x, y)$.

The given line is $\frac{x}{a}+\frac{y}{b}=1$ which can be written as $\frac{x}{a}+\frac{y}{b}-1=0$.

The foot of perpendicular is given as,

$\Rightarrow \frac{(x-0)}{\left(\frac{1}{a}\right)}=\frac{(y=0)}{\left(\frac{1}{b}\right)}=\frac{\left(\frac{1}{a} \times 0\right)+\left(\frac{1}{b} \times 0\right)-1}{\left(\frac{1}{a}\right)^{2}+\left(\frac{1}{b}\right)^{2}}$

$\Rightarrow a x=b y=\frac{-1}{\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)}$

Using the given relation $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}}$,

$\Rightarrow a x=b y=\frac{-1}{\left(\frac{1}{c^{2}}\right)}$

$\Rightarrow a x=b y=\left(-c^{2}\right)$

Equating the relations $a x=\left(-c^{2}\right)$ and $b y=\left(-c^{2}\right)$ gives

$x=\left(\frac{-c^{2}}{a}\right)$

and $y=\left(\frac{-c^{2}}{b}\right)$, respectively.

On squaring and adding,

$\Rightarrow x^{2}+y^{2}=\left(\frac{c^{4}}{a^{2}}\right)+\left(\frac{c^{4}}{b^{2}}\right)$

$\Rightarrow x^{2}+y^{2}=c^{4}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)$

$\Rightarrow x^{2}+y^{2}=c^{4}\left(\frac{1}{c^{2}}\right)$

$\Rightarrow x^{2}+y^{2}=c^{2}$

Ans: The given statement is false.

The given lines can be written are $a x+2 y+1=0$, $b x+3 y+1=0$ and $c x+4 y+1=0$. For these to be concurrent it must satisfy the determinant relation,

$\Rightarrow\left|\begin{array}{lll}

a & 2 & 1 \\

b & 3 & 1 \\

c & 4 & 1

\end{array}\right|=0$

Performing the row operations $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$,

$\Rightarrow\left|\begin{array}{ccc}

a & 2 & 1 \\

b-a & 1 & 0 \\

c-a & 2 & 0

\end{array}\right|=0$

Expanding the determinant along $C_{3}$,

$\Rightarrow 2(b-a)-(c-a)=0$

$\Rightarrow 2 b-a-c=0$

$\Rightarrow a+c=2 b$

Therefore, for the lines to be concurrent $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ must be in A.P.

Ans: The given statement is false.

The line joining the points  $ \left( {3,{\text{ }} - 4} \right) $  and  $ \left( { - 2,{\text{ }}6} \right) $  is,

$ \Rightarrow \left( {y - 6} \right) = \left( {\frac{{6 - \left( { - 4} \right)}}{{ - 2 - 3}}} \right)\left( {x - \left( { - 2} \right)} \right) $ 

$ \Rightarrow \left( {y - 6} \right) = \left( { - 2} \right)\left( {x + 2} \right) $ 

$ \Rightarrow y =  - 2x + 2 $   $ \left( {9,{\text{ }} - 18} \right) $ 

The slope of this line is  $  - 2 $ .

Similarly, the line joining the points  $ \left( { - 3,{\text{ }}6} \right) $  and  is,

$ \Rightarrow \left( {y - 6} \right) = \left( {\frac{{6 - \left( { - 18} \right)}}{{ - 3 - 9}}} \right)\left( {x - \left( { - 3} \right)} \right) $ 

$ \Rightarrow \left( {y - 6} \right) = \left( { - 2} \right)\left( {x + 3} \right) $ 

$ \Rightarrow y =  - 2x $  

The slope of this line is also  $  - 2 $ .

For the two lines to be perpendicular the product of their slopes must be  $  - 1 $ , but here the product is  $ \left( { - 2} \right) \times \left( { - 2} \right) = 4 $ .

Match the questions given under Column  $ {C_1} $  with their appropriate answers given under the column  $ {C_2} $  in Exercises  $ 57 $ to  $ 59 $ .

Ans:

(a) Let one of the required point is  $ \left( {a,{\text{ }}b} \right) $ . Since it lies on the line  $ x + 5y = 13 $ , so it will satisfy the line.

 $ \Rightarrow a + 5b = 13 $                       ……  $ \left( 1 \right) $ 

The distance of the point  $ \left( {a,{\text{ }}b} \right) $  from the line  $ 12x - 5y + 26 = 0 $  is  $ 2 $  units,

$ \Rightarrow \left| {\frac{{\left( {12 \times a} \right) + \left( {\left( { - 5} \right) \times b} \right) + 26}}{{\sqrt {{{\left( {12} \right)}^2} + {{\left( { - 5} \right)}^2}} }}} \right| = 2 $ 

$ \Rightarrow \left| {12a - 5b + 26} \right| = 26 $ 

$ \Rightarrow \left( {12a - 5b + 26} \right) =  \pm {\text{ }}26 $  

Considering the positive sign and solving equation  $ \left( 1 \right) $  and  $ \left( {12a - 5b + 26} \right) = 26 $  gives the point  $ \left( {a,{\text{ }}b} \right) = \left( {1,{\text{ }}\frac{{12}}{5}} \right) $ .

Considering the negative sign and solving equation  $ \left( 1 \right) $  and  $ \left( {12a - 5b + 26} \right) =  - 26 $  gives the point  $ \left( {a,{\text{ }}b} \right) = \left( { - 3,{\text{ }}\frac{{16}}{5}} \right) $ .

(b) Let one of the required point is  $ \left( {c,{\text{ }}d} \right) $ . Since it lies on the line  $ x + y = 4 $ , so it will satisfy the line.

 $ \Rightarrow c + d = 4 $                       ……  $ \left( 1 \right) $ 

The distance of the point  $ \left( {c,{\text{ }}d} \right) $  from the line  $ 4x + 3y - 10 = 0 $  is  $ 1 $  unit,

$ \Rightarrow \left| {\frac{{\left( {4 \times c} \right) + \left( {3 \times d} \right) - 10}}{{\sqrt {{{\left( 4 \right)}^2} + {{\left( 3 \right)}^2}} }}} \right| = 1 $ 

$ \Rightarrow \left| {4c + 3d - 10} \right| = 5 $ 

$ \Rightarrow \left( {4c + 3d - 10} \right) =  \pm {\text{ }}5 $  

Considering the positive sign and solving equation  $ \left( 1 \right) $  and  $ \left( {4c + 3d - 10} \right) = 5 $  gives the point  $ \left( {c,{\text{ }}d} \right) = \left( {3,{\text{ }}1} \right) $ .

Considering the negative sign and solving equation  $ \left( 1 \right) $  and  $ \left( {4c + 3d - 10} \right) =  - 5 $  gives the point  $ \left( {c,{\text{ }}d} \right) = \left( { - 7,{\text{ }}11} \right) $ .

(c) Here the points  $ P $  and  $ Q $  trisect the line  $ AB $ . Let the points be  $ P\left( {{x_1},{\text{ }}{y_1}} \right) $ and  $ Q\left( {{x_2},{\text{ }}{y_2}} \right) $ .

For point  $ P $ ,  $ AP:PB = m:n = 1:2 $ . The section formula states that if a point  $ \left( {x,y} \right) $  divides a line segments joining the two points  $ \left( {{x_1},{y_2}} \right) $  and  $ \left( {{x_2},{y_2}} \right) $  in the ratio  $ m:n $  then,

$ \left( {x,{\text{ }}y} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},{\text{ }}\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right) $ 

Using this formula for the points  $ A $ ,  $ P $  and  $ B $ ,

$ \Rightarrow \left( {{x_1},{\text{ }}{y_1}} \right) = \left( {\frac{{\left( {3 \times 1} \right) + \left( {2 \times \left( { - 2} \right)} \right)}}{{1 + 2}},{\text{ }}\frac{{\left( {1 \times 1} \right) + \left( {5 \times 2} \right)}}{{1 + 2}}} \right) $ 

$ \Rightarrow \left( {{x_1},{\text{ }}{y_1}} \right) = \left( { - \frac{1}{3},{\text{ }}\frac{{11}}{3}} \right) $  

Now,  $ P $  is the midpoint of  $ AQ $ . The midpoint formula is a special case of the section formula with  $ m:n = 1:1 $ .

Using the midpoint formula for  $ A $ ,  $ P $  and  $ Q $ , $ \Rightarrow \left( { - \frac{1}{3},{\text{ }}\frac{{11}}{3}} \right) = \left( {\frac{{ - 2 + {x_2}}}{2},{\text{ }}\frac{{5 + {y_2}}}{2}} \right) $ 

Solving for the coordinates of  $ Q $  by comparison gives  $ \left( {{x_2},{\text{ }}{y_2}} \right) = \left( {\frac{4}{3},{\text{ }}\frac{7}{3}} \right) $ .

Hence, the correct match is (a) – (iii), (b) – (i), (c) – (ii).

Ans: Simplifying the line  $ \left( {2x + 3y + 4} \right) + \lambda \left( {6x - y + 12} \right) = 0 $ 

gives  $ \left( {2 + 6\lambda } \right)x + \left( {3 - \lambda } \right)y + \left( {4 + 12\lambda } \right) = 0 $ .

(a) For the line  $ \left( {2 + 6\lambda } \right)x + \left( {3 - \lambda } \right)y + \left( {4 + 12\lambda } \right) = 0 $  to be parallel to the  $ y -  $  axis, the coefficient of  $ y -  $  coordinate must be  $ 0 $ .

$ \Rightarrow \left( {3 - \lambda } \right) = 0 $ 

$ \Rightarrow \lambda  = 3 $  

(b) The slope of the line  $ 7x + y - 4 = 0 $ is  $  - 7 $ , therefore the slope of the line  $ \left( {2 + 6\lambda } \right)x + \left( {3 - \lambda } \right)y + \left( {4 + 12\lambda } \right) = 0 $  must be  $ \frac{1}{7} $  so that the two lines are perpendicular.

$ \Rightarrow  - \frac{{\left( {2 + 6\lambda } \right)}}{{\left( {3 - \lambda } \right)}} = \frac{1}{7} $ 

$ \Rightarrow  - 14 - 42\lambda  = 3 - \lambda  $ 

$ \Rightarrow \lambda  =  - \frac{{17}}{{41}} $  

(c) If the point  $ \left( {1,{\text{ }}2} \right) $  passes through the line  $ \left( {2 + 6\lambda } \right)x + \left( {3 - \lambda } \right)y + \left( {4 + 12\lambda } \right) = 0 $  then it must satisfy the equation. On substitution,

$ \Rightarrow \left( {2 + 6\lambda } \right) + \left( {6 - 2\lambda } \right) + \left( {4 + 12\lambda } \right) = 0 $ 

$ \Rightarrow 16\lambda  =  - 12 $ 

$ \Rightarrow \lambda  =  - \frac{3}{4} $  

(d) For the line  $ \left( {2 + 6\lambda } \right)x + \left( {3 - \lambda } \right)y + \left( {4 + 12\lambda } \right) = 0 $  to be parallel to the  $ x -  $  axis, the coefficient of  $ x -  $  coordinate must be  $ 0 $ .

$ \Rightarrow \left( {2 + 6\lambda } \right) = 0 $ 

$ \Rightarrow \lambda  =  - \frac{1}{3} $  

Hence, the correct match is (a) – (iv), (b) – (iii), (c) – (i), (d) – (ii).

Ans: The point of intersection of the lines  $ 2x - 3y = 0 $  and  $ 4x - 5y = 2 $  is obtained by solving the equations algebraically. So the intersection point is  $ \left( {3,{\text{ }}2} \right) $ .

(a) The equation of the line passing through the points  $ \left( {2,{\text{ }}1} \right) $  and  $ \left( {3,{\text{ }}2} \right) $  is,

$ \Rightarrow \left( {y - 1} \right) = \left( {\frac{{2 - 1}}{{3 - 2}}} \right)\left( {x - 2} \right) $ 

$ \Rightarrow \left( {y - 1} \right) = \left( {x - 2} \right) $ 

$ \Rightarrow x - y - 1 = 0 $  

(b) The slope of the line  $ x + 2y + 1 = 0 $ is  $ \left( { - \frac{1}{2}} \right) $ , so the slope of the line perpendicular to this line will be  $ 2 $ . Now, the line passes through the point  $ \left( {3,{\text{ }}2} \right) $  so its equation is,

$ \Rightarrow \left( {y - 2} \right) = 2\left( {x - 3} \right) $ 

$ \Rightarrow \left( {y - 2} \right) = \left( {2x - 6} \right) $ 

$ \Rightarrow 2x - y = 4 $  

(c) The slope of the line  $ 3x - 4y + 5 = 0 $ is  $ \left( {\frac{3}{4}} \right) $ , so the slope of the line parallel to this line will also be  $ \left( {\frac{3}{4}} \right) $ . Now, the line passes through the point  $ \left( {3,{\text{ }}2} \right) $  so its equation is,

$ \Rightarrow \left( {y - 2} \right) = \left( {\frac{3}{4}} \right)\left( {x - 3} \right) $ 

$ \Rightarrow \left( {4y - 8} \right) = \left( {3x - 9} \right) $ 

$ \Rightarrow 3x - 4y - 1 = 0 $  

(d) The lines which are equally inclined to the axes have their slope values equal to  $ \tan {45^ \circ } = 1 $ or  $ \tan {135^ \circ } =  - 1 $ . Now, , the line passes through the point  $ \left( {3,{\text{ }}2} \right) $  so its equation is,

Considering the positive slope,

$ \Rightarrow \left( {y - 2} \right) = \left( 1 \right)\left( {x - 3} \right) $ 

$ \Rightarrow x - y - 1 = 0 $  

Considering the negative slope,

$ \Rightarrow \left( {y - 2} \right) = \left( { - 1} \right)\left( {x - 3} \right) $ 

$ \Rightarrow x + y - 5 = 0 $  

Hence, the correct match is (a) – (iii), (b) – (i), (c) – (iv), (d) – (ii) or (iii).