Question 1: Find the angles between each of the following pairs of straight lines:

(i) 3 x+y+12 = 0 and x+2y-1=0

(ii) 3x-y+5=0 and x -3y+1=0

(iii) 3x+4y-7 =0 and 4x-3y+5=0

(iv) x-4y=3 and 6x-y=11

v) (m^2-mn)y (mn+n^2)x + n^3 and (mn+m^2)y = (mn-n^2)x + m^3

Answer:

(i)     Let m_1 and m_2 be the slope of the straight lines 3 x+y+12 = 0 and x+2y-1=0 respectively.

Then m_1 = -3 and m_2 = \frac{-1}{2}

Let \theta   be the angle between the lines. Then,

\tan \theta = \Big| \frac{m_1 - m_2}{1 + m_1 m_2} \Big|  =  \Big| \frac{(-3) - (\frac{-1}{2}) }{1 + (-3) (\frac{-1}{2})} \Big| = \Big| \frac{-6+1}{2+3} \Big| = 1

\Rightarrow \theta = 45^{\circ}

Thus, the acute angle between the lines is of 45^{\circ}

(ii)     Let m_1 and m_2 be the slope of the straight lines 3x-y+5=0 and x -3y+1=0 respectively.

Then m_1 = 3 and m_2 = \frac{1}{3}

Let \theta   be the angle between the lines. Then,

\tan \theta = \Big| \frac{m_1 - m_2}{1 + m_1 m_2} \Big|  =  \Big| \frac{(3) - (\frac{1}{3}) }{1 + (3) (\frac{1}{3})} \Big| = \Big| \frac{9-1}{3+3} \Big| = \frac{4}{3}

\Rightarrow \theta = \tan^{-1} \frac{4}{3}

Thus, the acute angle between the lines is of \tan^{-1} \frac{4}{3}

(iii)    Let m_1 and m_2 be the slope of the straight lines 3x+4y-7 =0 and 4x-3y+5=0 respectively.

Then m_1 = \frac{-3}{4} and m_2 = \frac{4}{3}

Since m_1 \times m_2 = \frac{-3}{4} \times \frac{4}{3} = -1

Therefore the given lines are perpendicular.

Thus, the acute angle between the lines is of 90^{\circ}

(iv)    Let m_1 and m_2 be the slope of the straight lines x-4y=3 and 6x-y=11 respectively.

Then m_1 = \frac{1}{4} and m_2 = 6

Let \theta   be the angle between the lines. Then,

\tan \theta = \Big| \frac{m_1 - m_2}{1 + m_1 m_2} \Big|  =  \Big| \frac{(\frac{1}{4}) - 6 }{1 + (\frac{1}{4}) (6)} \Big| = \Big| \frac{1-24}{4+6} \Big| = \frac{23}{10}

\Rightarrow \theta = \tan^{-1} \frac{23}{10}

Thus, the acute angle between the lines is of \tan^{-1} \frac{23}{10}

v)      (m^2-mn)y (mn+n^2)x + n^3 and (mn+m^2)y = (mn-n^2)x + m^3

Then m_1 = \frac{mn+n^2}{m^2-mn} and m_2 = \frac{mn-n^2}{mn+m^2}

Let \theta   be the angle between the lines. Then,

\tan \theta = \Bigg| \frac{m_1 - m_2}{1 + m_1 m_2} \Bigg|  =  \Bigg| \frac{\frac{mn+n^2}{m^2-mn} - \frac{mn-n^2}{mn+m^2} }{1 + \frac{mn+n^2}{m^2-mn} \cdot \frac{mn-n^2}{mn+m^2} } \Bigg|

= \Big| \frac{m^2n^2+m^3n + mn^3 + m^2n^2-m^3n+m^2n^2+m^2n^2-mn^3}{m^3n+m^4-m^2n^2-m^3n+m^2n^2-mn^3+mn^3-n^4} \Big|

= \Big| \frac{4m^2n^2}{m^4-n^4} \Big|

\Rightarrow \theta = \tan^{-1} \Big( \frac{4m^2n^2}{m^4-n^4} \Big)

Thus, the acute angle between the lines is of \tan^{-1} \Big( \frac{4m^2n^2}{m^4-n^4} \Big)

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Question 2: Find the acute angle between the lines 2 x - y + 3 = 0 and x + y + 2 = 0 .

Answer:

Let m_1 and m_2 be the slope of the straight lines 2 x - y + 3 = 0 and x + y + 2 = 0 respectively.

Then m_1 = 2 and m_2 = -1

Let \theta   be the angle between the lines. Then,

\tan \theta = \Big| \frac{m_1 - m_2}{1 + m_1 m_2} \Big|  =  \Big| \frac{(2) - (-1) }{1 + (2) (-1)} \Big| = \Big| \frac{2+1}{1-2} \Big| = 3

\Rightarrow \theta = \tan^{-1} (3)

Thus, the acute angle between the lines is of \tan^{-1} (3)

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Question 3: Prove that the points (2, - 1), (0,2), (2,3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

Answer:

Let the vertices are  A(2, - 1), B(0,2), C(2,3) and D(4, 0)

Slope of AB = \frac{2-(-1)}{0-2} = \frac{-3}{2}

Slope of BC = \frac{3-2}{0-2} = \frac{1}{2}

Slope of CD = \frac{0-3}{4-2} = \frac{-3}{2}

Slope of DA = \frac{-1-0}{2-4} = \frac{1}{2}

Therefore AB \parallel CD and BC \parallel DA

Hence ABCD is a parallelogram.

Let m_1 and m_2 be the slope of the straight lines AC and BD respectively.

Then m_1 = \frac{3-(-1)}{2-2} = \infty     (AC is \parallel to y-axis )

m_2 = \frac{0-2}{4-0} = \frac{-1}{2}

\tan \theta = \Big| \frac{m_1 - m_2}{1 + m_1 m_2} \Big| =  \Big| \frac{1 - \frac{m_2}{m_1} }{\frac{1}{m_1} + m_2} \Big| = \Big| \frac{1-0}{0-\frac{1}{2}} \Big| = 2

\Rightarrow \theta = \tan^{-1} (2)

Thus, the acute angle between the diagonals is of \tan^{-1} (2)

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Question 4: Find the angle between the line joining the points (2,0), (0,3) and the line x + y =1 .

Answer:

Let the points be A(2,0), B(0,3)

Slope of AB  ( m_1)  = \frac{3-0}{0-2} = \frac{-3}{2}

Slope of x + y =1   m_2 = -1

Let \theta   be the angle between the lines. Then,

\tan \theta = \Big| \frac{m_1 - m_2}{1 + m_1 m_2} \Big|  =  \Big| \frac{(\frac{-3}{2}) - (-1) }{1 + (\frac{-3}{2}) (-1)} \Big| = \Big| \frac{-3+2}{2+3} \Big| = \frac{1}{5}

\Rightarrow \theta = \tan^{-1} \Big( \frac{1}{5} \Big)

Thus, the acute angle between the lines is of \tan^{-1} \Big( \frac{1}{5} \Big)

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Question 5: If \theta is the angle which the straight line joining the points (x_1, y_1) and (x_2, y_2) subtends at the origin, prove that \tan \theta = \frac{x_2y_1 - x_1y_2}{x_1x_2 + y_1y_2}   and           \cos \theta = \frac{x_1x_2 + y_1y_2}{ \sqrt{{x_1}^2 + {y_1}^2}  \sqrt{{x_2}^2 + {y_2}^2} }

Answer:

Let the two points be A ( x_1, y_1) and B ( x_2, y_2)

Slope of OA = m_1 = \frac{y_1-0}{x_1-0} = \frac{y_1}{x_1}

Slope of OB = m_2 = \frac{y_2-0}{x_2-0} = \frac{y_2}{x_2}

\therefore \tan \theta = \Bigg|  \frac{\frac{y_1}{x_1} - \frac{y_2}{x_2}}{1 + \frac{y_1}{x_1} \cdot \frac{y_2}{x_2}}   \Bigg|  = \Big|  \frac{y_1x_2-y_2x_1}{x_1x_2+y_1y_2} \Big|

We know \cos \theta = \sqrt{ \frac{1}{1 + \tan^2 \theta} }

= \frac{x_1x_2+y_1y_2}{ \sqrt{( x_1x_2+y_1y_2)^2 + ( x_2y_1-x_1y_2)^2}  }

= \frac{x_1x_2+y_1y_2}{ \sqrt{ {x_1}^2{x_2}^2 + {y_1}^2 {y_1}^2 + 2 x_1x_2y_1y_2 +  {x_2}^2{y_1}^2 + {x_1}^2 {y_2}^2 - 2 x_1x_2y_1y_2  }}

= \frac{x_1x_2+y_1y_2}{ \sqrt{ {x_1}^2 ( {x_2}^2+ {y_2}^2) + {y_1}^2 ( {x_2}^2+ {y_2}^2)  }}

= \frac{x_1x_2+y_1y_2}{ \sqrt{ ({x_1}^2 + {y_1}^2) ( {x_2}^2+ {y_2}^2)   }}

= \frac{x_1x_2+y_1y_2}{ \sqrt{ ({x_1}^2 + {y_1}^2)} \sqrt{( {x_2}^2+ {y_2}^2)   }}

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Question 6: Prove that the straight lines (a+ b)x+(a-b)y=2ab,(a-b)x + (a+b) y=2ab and x + y =0 form an isosceles triangle whose vertical angle is 2 \tan^{-1} \Big( \frac{a}{b} \Big)

Answer:

Given Lines:

Line i)   : (a+ b)x+(a-b)y=2ab       \Rightarrow  \text{Slope   } = m_1 = \frac{-(a+b)}{(a-b)}

Line ii)  : (a-b)x + (a+b) y=2ab         \Rightarrow  \text{Slope   } = m_2 = \frac{-(a-b)}{(a+b)}

Line iii) : x + y =0       \Rightarrow  \text{Slope} = m_3 = -1

Let \theta_1 is the angle between line i) and line ii)

\tan \theta_1 = \Bigg|  \frac{\frac{-(a+b)}{(a-b)}+\frac{-(a-b)}{(a+b)} }{1 + \frac{(a+b)}{(a-b)} \cdot  \frac{(a-b)}{(a+b)} } \Bigg| = \Big| \frac{2ab}{b^2-a^2}   \Big| = \Big|  \frac{2( \frac{a}{b})}{ 1 - ( \frac{a}{b})^2} \Big|

\tan \theta_1 = \tan \Big( 2 \tan^{-1} \Big( \frac{a}{b} \Big ) \Big)

\Rightarrow \theta_1 = 2 \tan^{-1} \Big( \frac{a}{b} \Big )

Let \theta_2 is the angle between line ii) and line iii)

\tan \theta_2 = \Bigg| \frac{\frac{-(a-b)}{(a+b) + 1}}{1 + \frac{(a-b)}{(a+b)}}     \Bigg| =    \Big| \frac{-a+b+a+b}{a+b+a-b}   \Big|    =    \Big| \frac{b}{a}   \Big|

\Rightarrow \theta_2 = \tan^{-1} \Big( \frac{b}{a} \Big)

Let \theta_3 is the angle between line iii) and line i)

\tan \theta_2 = \Bigg| \frac{\frac{-1 + (a+b)}{(a-b)}}{1 + (-1) \frac{-(a+b)}{(a-b)} }    \Bigg| =    \Big|  \frac{-a+b+a+b}{a-b+a+b}   \Big|    =    \Big| \frac{b}{a}   \Big|

Therefore the triangle is an isosceles as \theta_2 = \theta_3

The vertical angle is \theta_1 = 2 \tan^{-1} \Big( \frac{a}{b} \Big )

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Question 7: Find the angle between the lines x = a and by + c =0 .

Answer:

Given: Line i) : x = a   \Rightarrow \text{Slope   } = m_1 = \frac{1}{0} . This line is \parallel to y-axis.

Line ii) by + c =0   \Rightarrow \text{Slope   } = m_2 = 0 . This line is \parallel to x-axis.

\therefore \tan \theta = \Bigg| \frac{ \frac{1}{0} - 0}{1 + \frac{1}{0} \cdot 0} \Bigg| = \Big| \frac{1}{0} \Big| = \infty

\therefore \theta = \tan^{-1} ( \infty) = 90^{\circ}

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Question 8: Find the tangent of the angle between the lines which have intercepts 3,4 and 1, 8 on the axes respectively.

Answer:

Equation of the line with intercepts 3, 4 :

\frac{x}{3} + \frac{y}{4} = 1 \Rightarrow 4x + 3y = 12      … … … … … i)

Equation of the line with intercepts 1, 8 :

\frac{x}{1} + \frac{y}{8} = 1 \Rightarrow 8x + y = 8      … … … … … ii)

Slope of line i) m_1 = \frac{-4}{3}

Slope of line ii) m_2 = -8

\therefore \tan \theta = \Bigg| \frac{ \frac{-4}{3} - (-8)}{ 1 + ( \frac{-4}{3}) ( -8) } \Bigg|  = \Big| \frac{-4+24}{3+32}   \Big|  = \frac{20}{35} = \frac{4}{7}

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Question 9: Show that the line a^2 x + ay + 1= 0 is perpendicular to the line x-ay=1 for all non-zero real values of a .

Answer:

Line i) : a^2 x + ay + 1 = 0 \Rightarrow \text{Slope   } m_1 = \frac{-a^2}{a} =-a

Line ii) : x-ay=1 \Rightarrow \text{Slope   } m_2 = \frac{1}{a}

Now, m_1 \times m_2 = (-a) \times \Big( \frac{1}{a} \Big) = - 1

Therefore the two lines are perpendicular to each other.

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Question 10: Show that the tangent of an angle between the lines \frac{x}{a} + \frac{y}{b} = 1 and \frac{x}{a} - \frac{y}{b} = 1 is \frac{2ab}{a^2- b^2}

Answer:

Given lines:

\frac{x}{a} + \frac{y}{b} = 1 \Rightarrow bx + ay = ab      … … … … … i)

\frac{x}{a} - \frac{y}{b} = 1 \Rightarrow bx -a y = ab      … … … … … ii)

Slope of line i) m_1 = \frac{-b}{a}

Slope of line ii) m_2 = \frac{b}{a}

\therefore \tan \theta = \Bigg| \frac{ \frac{-b}{a} - \frac{b}{a}}{ 1 + ( \frac{b}{a}) ( -8) } \Bigg|  = \Big| \frac{-2ab}{a^2-b^2} \Big|= \frac{2ab}{a^2-b^2}