Chapter 10: Straight Lines

Ans: Given: $ x + y - 2 = 0 $ 

Let  $ ABC $  be the equilateral triangle with vertex  $ A(h,k) $ .

The centroid is  $ G(0,0) $ .

Slope of line  $ BC $  i.e.  $ x + y - 2 = 0 $ is  $  - 1 $ .

As $ AG \bot BC $ ,

The slope of  $ AG $  is  $ \frac{k}{h} = 1 $ or  $ h = k $ .

The distance of  $ (0,0) $ from  $ BC = \frac{{|0 + 0 - 2|}}{{\sqrt {{1^2} + {1^2}} }} = \sqrt 2  $ .

Distance of  $ A $  from  $ BC = 3\sqrt 2  = \frac{{|h + k - 2|}}{{\sqrt {{1^2} + {1^2}} }} $ 

 $ \Rightarrow |h + k - 2| = 6 $ 

 $ \Rightarrow h + k - 8 = 0 $ or $ h + k + 4 = 0 $ 

 $ \Rightarrow h + h - 8 = 0 $ or $ h + h + 4 = 0 $ 

 $ \Rightarrow h = 4 $ or $ h =  - 2 $ 

Hence, the vertex is  $ ( - 2, - 2) $ 

Correct Option:C

Ans: Given: $ {\text{a, b, c}} $  are in A.P.

 $ {\text{a, b, c}} $ are in A.P.

So,  $ b = \frac{{a + c}}{2} $  i.e.  $ a - 2b + c = 0 $ 

Compare  $ a - 2b + c = 0 $  and line $ ax + by + c = 0 $ ,

 $ \Rightarrow x = 1,y =  - 2 $ 

Hence, the point  $ (1, - 2) $  lies on the line.

Ans: Given: $ (1, - 2) $ 

The line which cuts equal intercepts from the axes is  $ \frac{x}{a} + \frac{y}{a} = 1 $ .

As this line passes through  $ (1, - 2), $ 

 $ \Rightarrow 1 - 2 = a $ or  $ a =  - 1 $ .

Thus, the required equation of line is:  $ x + y =  - 1 $ 

 $ \Rightarrow x + y + 1 = 0 $ 

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

Ans: Given: Line passes through  $ (3,2) $ .

Slope of  $ x - 2y = 3 $ is  $ \frac{1}{2} $ . 

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

 $ \Rightarrow \tan {45^\circ } = \left| {\frac{{m - \frac{1}{2}}}{{1 + \frac{1}{2}m}}} \right| $ 

 $ \Rightarrow 1 =  \pm \frac{{2m - 1}}{{2 + m}} $ 

 $ \Rightarrow 2m - 1 = 2 + m{\text{ or }}1 - 2m = 2 + m $ 

 $ \Rightarrow m = 3{\text{ or }}m =  - \frac{1}{3} $ 

The line also passes through $ (3,2) $ .

The equation of the line is:  $ y - 2 = 3(x - 3) $  or  $ y - 2 =  - \frac{1}{3}(x - 3) $ 

Hence, equations of lines are:  $ 3x - y - 7 = 0,{\text{  }}x + 3y - 9 = 0 $ .

Ans: Given: $ (3,4) $ and  $ (2, - 6) $ .

Given line is  $ 3x - 4y - 8 = 0 $ .

For the point  $ (3,4): $  $ 3(3) - 4(4) - 8 =  - 15 < 0 $ 

For the point  $ (2, - 6): $  $ 3(2) - 4( - 6) - 8 = 22 > 0 $ 

Hence,  $ (3,4) $  and  $ (2, - 6) $  lie on opposite side of  $ 3x - 4y - 8 = 0 $ .

Ans: Given: $ (3, - 2) $ .

Let the moving point be  $ P(h,k) $ . 

Given point is  $ {\text{A}}(3, - 2) $ .

 $ \Rightarrow A{P^2} = {(h - 3)^2} + {(k + 2)^2} = d_1^2 $ 

The distance of  $ (h,k) $  from  $ 5x - 12y - 3 = 0 $  is:

 $ \Rightarrow {d_2} = \left| {\frac{{5h - 12k - 3}}{{\sqrt {25 + 144} }}} \right| $ 

 $ \Rightarrow {d_2} = \left| {\frac{{5h - 12k - 3}}{{13}}} \right| $ 

Given that,  $ d_1^2 = {d_2} $ 

 $ \Rightarrow {(h - 3)^2} + {(k + 2)^2} = \frac{{5h - 12k - 3}}{{13}}\quad  $ 

 $ \Rightarrow {h^2} - 6h + 9 + {k^2} + 4k + 4 = \frac{{5h - 12k - 3}}{{13}} $ 

 $ \Rightarrow 13{h^2} + 13{k^2} - 78h + 52k + 169 = 5h - 12k - 3 $ 

 $ \Rightarrow 13{h^2} + 13{k^2} - 83h + 64k + 172 = 0 $ 

Hence, locus is:  $ 13{x^2} + 13{y^2} - 83x + 64y + 172 = 0 $ 

Ans: Given: $ x\sin \theta  + y\cos \theta  = p $ .

 $ x\sin \theta  + y\cos \theta  = p $ meets axes at points $ A\left( {\frac{p}{{\sin \theta }},0} \right) $  and  $ B\left( {0,\frac{p}{{\cos \theta }}} \right) $ .

Let  $ P(h,k) $  be the mid-point of  $ AB. $ 

 $ \Rightarrow h = \frac{p}{{2\sin \theta }} $ and $ k = \frac{p}{{2\cos \theta }} $ 

 $ \Rightarrow \sin \theta  = \frac{p}{{2h}} $ and $ \cos \theta  = \frac{p}{{2k}} $ 

By squaring and adding,

 $ \Rightarrow {\sin ^2}\theta  + {\cos ^2}\theta  = \frac{{{p^2}}}{{4{h^2}}} + \frac{{{p^2}}}{{4{k^2}}} $ 

 $ \Rightarrow 1 = \frac{{{p^2}}}{{4{x^2}}} + \frac{{{p^2}}}{{4{y^2}}} $ 

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

Hence, the locus is  $ 4{x^2}{y^2} = {p^2}\left( {{x^2} + {y^2}} \right) $ .

State whether the Statements in Exercises  $ 48 $  to  $ 56 $  True or False? Justify.

Ans:The given statement is true.

Let us assume an equilateral triangle with vertices  $ A\left( {{x_1},{\text{ }}{y_1}} \right) $ ,  $ B\left( {{x_2},{\text{ }}{y_2}} \right) $  and  $ C\left( {{x_3},{\text{ }}{y_3}} \right) $ . All the angles of an equilateral triangle are  $ {60^ \circ } $ . Assuming the slope of line  $ AB $  as  $ {m_1} $  and the slope of line  $ BC $ as  $ {m_2} $ .

 $ \Rightarrow \tan {60^ \circ } = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right| $ 

 $ \Rightarrow \sqrt 3  = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right| $                       ……  $ \left( 1 \right) $ 

Here  $ {m_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} $ and  $ {m_2} = \frac{{{y_3} - {y_2}}}{{{x_3} - {x_2}}} $ . Since all the coordinates are integers so  $ {m_1} $  and  $ {m_2} $  are rational numbers.

Therefore, in expression  $ \left( 1 \right) $ , the R.H.S is a rational number and the L.H.S is an irrational number. This results in rational number  $  =  $  irrational numbers which are not possible.

Ans: The given statement is false.

Using the distance formula, the length  $ AB $ 

$ = \sqrt {{{\left( {0 - \left( { - 2} \right)} \right)}^2} + {{\left( {5 - 1} \right)}^2}}  $ 

$ = \sqrt {4 + 16}  $ 

$ = 2\sqrt 5 {\text{ units}} $  

Similarly, the length  $ BC $ 

$ = \sqrt {{{\left( { - 1 - 0} \right)}^2} + {{\left( {2 - 5} \right)}^2}}  $ 

$ = \sqrt {1 + 9}  $ 

$ = \sqrt {10} {\text{ units}} $  

And, the length  $ AC $ 

$ = \sqrt {{{\left( { - 2 - \left( { - 1} \right)} \right)}^2} + {{\left( {1 - 2} \right)}^2}}  $ 

$ = \sqrt {1 + 1}  $ 

$ = \sqrt 2 {\text{ units}} $  

Clearly the sum of any two line segments is not equal to the third one in any case, so the points are not collinear.

Ans:The given statement is false.

The slope of the given line  $ x\sec \theta  + y\cos {\text{ec}}\theta  = a $  is,

$ \Rightarrow {m_1} =  - \frac{{\sec \theta }}{{\cos {\text{ec}}\theta }} $ 

$ \Rightarrow {m_1} =  - \frac{{\left( {\frac{1}{{\cos \theta }}} \right)}}{{\left( {\frac{1}{{\sin \theta }}} \right)}} $ 

$ \Rightarrow {m_1} =  - \left( {\frac{{\sin \theta }}{{\cos \theta }}} \right) $  

Therefore, the slope of the required line which is perpendicular to the given line is,

$ \Rightarrow {m_2} = \frac{{ - 1}}{{{m_1}}} $ 

$ \Rightarrow {m_2} = \frac{{ - 1}}{{ - \left( {\frac{{\sin \theta }}{{\cos \theta }}} \right)}} $  

$ \Rightarrow {m_2} = \frac{{\cos \theta }}{{\sin \theta }} $  

So the equation of the line passing through the point  $ \left( {a{{\cos }^3}\theta ,{\text{ }}a{{\sin }^3}\theta } \right) $  and having slope  $ {m_2} $  is,

$ \Rightarrow \left( {y - a{{\sin }^3}\theta } \right) = {m_2}\left( {x - a{{\cos }^3}\theta } \right) $ 

$ \Rightarrow \left( {y - a{{\sin }^3}\theta } \right) = \frac{{\cos \theta }}{{\sin \theta }}\left( {x - a{{\cos }^3}\theta } \right) $ 

$ \Rightarrow \sin \theta \left( {y - a{{\sin }^3}\theta } \right) = \cos \theta \left( {x - a{{\cos }^3}\theta } \right) $ 

$ \Rightarrow x\cos \theta  - y\sin \theta  = a{\cos ^4}\theta  - a{\sin ^4}\theta  $  

Using the algebraic identity  $ {m^2} - {n^2} = \left( {m + n} \right)\left( {m - n} \right) $  and the trigonometric identity  $ {\sin ^2}\theta  + {\cos ^2}\theta  = 1 $ ,

$ \Rightarrow x\cos \theta  - y\sin \theta  = a\left( {{{\cos }^2}\theta  - {{\sin }^2}\theta } \right)\left( {{{\cos }^2}\theta  + {{\sin }^2}\theta } \right) $ 

$ \Rightarrow x\cos \theta  - y\sin \theta  = a\left( {{{\cos }^2}\theta  - {{\sin }^2}\theta } \right) $  

Using the identity  $ {\cos ^2}\theta  - {\sin ^2}\theta  = \cos 2\theta  $ ,

$ \Rightarrow x\cos \theta  - y\sin \theta  = a\cos 2\theta  $