Chapter 10: Straight Lines

Ans: 

Given: $ y = mx + {c_1} $ and  $ y = mx + {c_2} $ .

Let  $ P\left( {{x_1},{y_1}} \right) $  be a point on $ y = mx + {c_1} $ .

The other line is:  $ y = mx + {c_2} $  i.e.  $ mx - y + {c_2} = 0 $ .

Distance of  $ P\left( {{x_1},{y_1}} \right) $  from this line is  $ d = \frac{{\left| {m{x_1} - {y_1} + {c_2}} \right|}}{{\sqrt {{m^2} + 1} }} $ .

 $ P\left( {{x_1},{y_1}} \right) $ lies on  $ y = mx + {c_1} $ ,

 $ \Rightarrow {y_1} = m{x_1} + {c_1} $ 

 $ \Rightarrow m{x_1} - {y_1} =  - {c_1} $ 

 $ \Rightarrow d = \frac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {{m^2} + 1} }} $ 

Hence, distance is  $ \frac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {{m^2} + 1} }} $ .

Correct Option: B

Ans: 

Given:  $ y = 3x + 4 $ and  $ (2,3) $ .

Let the foot of perpendicular from  $ P(2,3) $  on  $ 3x - y + 4 = 0 $  be $ M(h,k) $ .

 $ M(h,k) $ lies on  $ y = 3x + 4 $ .

 $ \Rightarrow 3h - k + 4 = 0{\text{              }}...(1) $ 

The slope of the given line is  $ 3 $  .

Slope of the line joining  $ P(2,3) $  and  $ M(h,k) $  is  $  - \frac{1}{3} = \frac{{k - 3}}{{h - 2}} $  or  $ h + 3k - 11 = 0{\text{             }}...{\text{(2)}} $ 

Solve  $ {\text{(1)}} $ and  $ {\text{(2)}} $ , 

 $ \Rightarrow (h,k) = \left( { - \frac{1}{{10}},\frac{{37}}{{10}}} \right) $ 

Hence, the coordinates of foot of perpendicular is  $ \left( { - \frac{1}{{10}},\frac{{37}}{{10}}} \right) $ .

Correct Option: B

Ans: 

Given: $ (3,2) $ 

As the middle point is  $ P(3,2) $ , the line meets the axes at  $ A(6,0) $  and  $ B(0,4) $ .

So, the equation of the line using intercept form is:  $ \frac{x}{6} + \frac{y}{4} = 1 $  i.e.  $ 2x + 3y = 12 $ .

Hence, the equation is  $ 2x + 3y = 12 $ . 

Correct Option: A

Ans: 

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

The required line is parallel to $ y = 3x - 1 $ .

Thus, slope of the line is  $ 3 $ .

The line also passes through  $ (1,2) $ .

So, the equation of line is:  $ y - 2 = 3(x - 1) $ 

Hence, the equation is  $ y - 2 = 3(x - 1) $ . 

Correct Option: C

Ans: 

Given: $ x = 0,{\text{ }}y = 0,{\text{ }}x = 1 $ and  $ y = 1 $ .

Using the equations  $ x = 0,{\text{ }}y = 0,{\text{ }}x = 1 $ and  $ y = 1 $ , the vertices of square are  $ A(0,0),{\text{ }}B(1,0),{\text{ }}C(1,1),{\text{ }}D(0,1) $ .

The equation of  $ AC $  is:  $ y - 0 = \frac{{1 - 0}}{{1 - 0}}(x - 0) $  i.e.  $ x - y = 0 $ .

The equation of  $ BD $  is:  $ y - 0 = \frac{{1 - 0}}{{0 - 1}}(x - 1) $  i.e.  $ x + y - 1 = 0 $ .

Hence, the diagonals are $ y = x,{\text{ }}y + x = 1 $ .

Correct Option: A

Ans: 

Given: A straight line is specified.

The general equation of a linear equation in two variables is  $ ax + by + c = 0 $ 

At least one of  $ a $  and  $ b $  should be non-zero. 

Assume $ a \ne 0 $ . 

Then, the equation of the line is:  $ x + \frac{b}{a}y + \frac{c}{a} = 0 $ 

Hence, two parameters should be known.

Correct Option: B

Ans: 

Given: $ (4,1) $  undergoes two successive transformations.

The reflection of  $ A(4,1) $  in  $ {\text{y}} = {\text{x}} $ is  $ B(1,4) $ .

Then, translation of  $ B $ through a distance  $ 2 $  units along the positive  $ x -  $ axis shifts  $ B $  to  $ C(1 + 2,4) $  i.e.  $ C(3,4) $ .

Hence, the final coordinates of point is  $ (3,4) $ .

Correct Option: B 

Ans: Given: $ 4x + 3y + 10 = 0,{\text{ }}5x - 12y + 26 = 0 $ and  $ 7x + 24y - 50 = 0 $ .

The distance of  $ (0,0) $  from  $ 4x + 3y + 10 = 0 $ is  $ \frac{{|0 + 0 + 10|}}{{\sqrt {{4^2} + {3^2}} }} = 2 $ .

The distance of  $ (0,0) $  from  $ 5x - 12y + 26 = 0 $ is  $ \frac{{|0 + 0 + 26|}}{{\sqrt {{5^2} + {{12}^2}} }} = 2 $ .

The distance of  $ (0,0) $  from  $ 7x + 24y - 50 = 0 $ is  $ \frac{{|0 + 0 - 50|}}{{\sqrt {{7^2} + {{24}^2}} }} = 2 $ .

The distance of all the three lines from  $ (0,0) $  is  $ 2 $ units.

Hence,  $ (0,0) $  is the required point.

Correct Option: C

Ans:

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

The slope of  $ 3x + y = 3 $ is  $  - 3 $ .

So, slope of a line perpendicular to it $  = \frac{1}{3} $ .

The equation of the required line is:  $ y - 2 = \frac{1}{3}(x - 2) $  i.e.  $ x - 3y + 4 = 0 $ .

Put  $ x = 0 $  to find  $ y -  $ intercept.

 $ \Rightarrow 0 - 3y + 4 = 0 $ 

 $ \Rightarrow y = \frac{4}{3} $ which is the $ y -  $ intercept.

Hence, the $ y -  $ intercept is  $ \frac{4}{3} $ .

Correct Option: D

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

Distance between  $ 3x + 4y + 5 = 0 $  and  $ 3x + 4y + 2 = 0 $  is  $ \frac{{|5 - 2|}}{{\sqrt {9 + 16} }} = \frac{3}{5} $ 

Distance between  $ 3x + 4y - 5 = 0 $  and  $ 3x + 4y + 2 = 0 $  is  $ \frac{{| - 5 - 2|}}{{\sqrt {9 + 16} }} = \frac{7}{5} $ 

Hence, the ratio is  $ \frac{3}{5}:\frac{7}{5} $  i.e.  $ 3:7 $ .

Correct Option: B