Note: We know that the equations of two lines passing through ( x_1, y_1)   and making and angle \alpha   with the given line y = mx + c   are

y - y_1 = \Big(  \frac{m \pm \tan \alpha }{1 \mp m \tan \alpha} \Big)  ( x - x_1)

Question 1: Find the equation of the straight lines passing through the origin and making an angle of 45^{\circ} with the straight line \sqrt{3} x + y =11 .

Answer:

We know that the equations of two lines passing through ( x_1, y_1)   and making and angle \alpha   with the given line y = mx + c   are

y - y_1 = \Big(  \frac{m \pm \tan \alpha }{1 \mp m \tan \alpha} \Big)  ( x - x_1)

Here x_1  = 0, \hspace{0.5cm} y_1 = 0 , \hspace{0.5cm} \alpha = 45^{\circ}  \hspace{0.5cm} m = -\sqrt{3}

Therefore the equation are:

y - 0 = \Big(  \frac{-\sqrt{3} + \tan 45^{\circ} }{1 + \sqrt{3} \tan 45^{\circ}} \Big)  ( x - 0)

\Rightarrow  y = \Big(  \frac{-\sqrt{3} + 1 }{1 + \sqrt{3} } \Big)  x

\Rightarrow  y = \Big(  \frac{-\sqrt{3} + 1 }{1 + \sqrt{3} } \times \frac{\sqrt{3} - 1}{\sqrt{3} -1} \Big)  x

\Rightarrow  y = \Big( \frac{3 +1  - 2\sqrt{2} }{3+1 } \Big)  x

\Rightarrow  y = ( \sqrt{3} - 2)   x      … … … … … i)

y - 0 = \Big(  \frac{-\sqrt{3} - \tan 45^{\circ} }{1 - \sqrt{3} \tan 45^{\circ}} \Big)  ( x - 0)

\Rightarrow  y = \Big(  \frac{-\sqrt{3} - 1 }{1 - \sqrt{3} } \Big)  x

\Rightarrow  y = \Big(  \frac{\sqrt{3} + 1 }{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} +1} \Big)  x

\Rightarrow  y = \Big( \frac{3 +1  + 2\sqrt{2} }{3-1 } \Big)  x

\Rightarrow  y = ( \sqrt{3} + 2)   x      … … … … … ii)

Therefore y = ( \sqrt{3} - 2)   x  \text{ and } y = ( \sqrt{3} + 2)   x are the two equations.

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Question 2: Find the equations to the straight lines which pass through the origin and are inclined at an angle of 75^{\circ} to the straight line x + y +\sqrt{3} (y -x) =a .

Answer:

Given equation:

x+ y + \sqrt{3}( y - x) = a      \Rightarrow ( \sqrt{3}+1) y = ( \sqrt{3}-1) x + a      \Rightarrow y = \frac{\sqrt{3}-1}{\sqrt{3}+1} x + \frac{a}{\sqrt{3}+1}

\therefore \text{ Slope } m = \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} = 2 - \sqrt{3}

Here x_1  = 0, \hspace{0.5cm} y_1 = 0 , \hspace{0.5cm} \alpha = 75^{\circ}  \hspace{0.5cm} m = 2-\sqrt{3}

Therefore the equations are:

y - 0 = \Big(  \frac{(2-\sqrt{3}) + \tan 75^{\circ} }{1 -( 2 -  \sqrt{3}) \tan 75^{\circ}} \Big)  ( x - 0)

\Rightarrow  y = \Big(  \frac{(2-\sqrt{3}) + ( 2 + \sqrt{3}) }{1 - (2 -  \sqrt{3})(2+\sqrt{3} } \Big)  x

\Rightarrow  y = \Big(  \frac{4 }{0 } \Big)  x

\Rightarrow  x = 0      … … … … … i)

y - 0 = \Big(  \frac{(2-\sqrt{3}) - \tan 75^{\circ} }{1 +( 2 -  \sqrt{3}) \tan 75^{\circ}} \Big)  ( x - 0)

\Rightarrow  y = \Big(  \frac{(2-\sqrt{3}) - ( 2 + \sqrt{3}) }{1 + (2 -  \sqrt{3})(2+\sqrt{3} } \Big)  x

\Rightarrow  y = \Big(  \frac{-2\sqrt{3} }{1+1 } \Big)  x

\Rightarrow  y + \sqrt{3} x = 0      … … … … … ii)

Therefore x = 0  \text{ and } y + \sqrt{3} x = 0 are the two equations.

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Question 3: Find the equations of the straight lines passing through (2, - 1) and making an angle of 45^{\circ} with the line 6x + 5 y - 8 = 0 .

Answer:

Given equation:

6x + 5 y - 8 = 0      \Rightarrow y = \frac{-6}{5} x + \frac{4}{3}      \therefore \text{ Slope } m = \frac{-6}{5}

Here x_1  = 2, \hspace{0.5cm} y_1 = 2 , \hspace{0.5cm} \alpha = 45^{\circ}  \hspace{0.5cm} m = \frac{-6}{5}

Therefore the equations are:

y - (-1) = \Big(  \frac{ \frac{-6}{5} + \tan 45^{\circ} }{1 - ( \frac{-6}{5}) \tan 45^{\circ}} \Big)  ( x - 2)

\Rightarrow y +1 = \Big(  \frac{ \frac{-6}{5} + 1 }{1 + ( \frac{6}{5}) } \Big)  ( x - 2)

\Rightarrow  y + 1 = \frac{-1}{11} ( x- 2)

\Rightarrow  x + 11y + 9 = 0      … … … … … i)

y - (-1) = \Big(  \frac{\frac{-6}{5} - \tan 45^{\circ} }{1 - ( \frac{6}{5}) \tan 45^{\circ}} \Big)  ( x - 2)

\Rightarrow y +1 = \Big(  \frac{\frac{-6}{5} - 1 }{1 - ( \frac{6}{5}) } \Big)  ( x - 2)

\Rightarrow  y + 1 = \frac{-11}{-1} ( x- 2)

\Rightarrow  11x-y-23 = 0      … … … … … ii)

Therefore x + 11y + 9 = 0  \text{ and } 11x-y-23 = 0 are the two equations.

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Question 4: Find the equations to the straight lines which pass through the point (h, k) and are inclined at angle \tan^{-1} m to the straight line y = mx + c .

Answer:

Here x_1  = h, \hspace{0.5cm} y_1 = k , \hspace{0.5cm} \alpha = \tan^{-1} m  \hspace{0.5cm} m = m

Therefore the equations are:

y - k = \Big(  \frac{ m + m }{1 - m^2 } \Big)  ( x - h)

\Rightarrow y - k = \Big(  \frac{ 2m }{1 - m^2 } \Big)  ( x - h)

\Rightarrow  (y-k)(1-m^2) = 2m ( x - h)

\Rightarrow   (1- m^2)y - 2mx = k( 1-m^2) - 2mh … … … … … i)

y - k = \Big(  \frac{ m - m }{1 - m^2 } \Big)  ( x - h)

\Rightarrow y - k = 0  ( x - h)

\Rightarrow  y = k      … … … … … ii)

Therefore (1- m^2)y - 2nx = k( 1-m^2) - 2mh  \text{ and } y = k are the two equations.

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Question 5: Find the equations to the straight lines passing through the point (2, 3) and inclined at an angle of 45^{\circ} to the line 3 x + y -5 = 0 .

Answer:

Given equation:

3x+y-5=0      \Rightarrow y = - 3x + 5      \therefore \text{ Slope } m = -3

Here x_1  = 2, \hspace{0.5cm} y_1 = 3 , \hspace{0.5cm} \alpha = 45^{\circ}  \hspace{0.5cm} m = -3

Therefore the equations are:

y - (3) = \Big(  \frac{ -3 + \tan 45^{\circ} }{1 + 3  \tan 45^{\circ}} \Big)  ( x - 2)

\Rightarrow y -3 = \Big(  \frac{ -3 + 1 }{1 + 3 } \Big)  ( x - 2)

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

\Rightarrow  x + 2y -8 = 0      … … … … … i)

y - (3) = \Big(  \frac{ -3 - \tan 45^{\circ} }{1 - 3  \tan 45^{\circ}} \Big)  ( x - 2)

\Rightarrow y -3 = \Big(  \frac{ -3 - 1 }{1 3 3 } \Big)  ( x - 2)

\Rightarrow  y -3 = 2 ( x- 2)

\Rightarrow  2x-y-1= 0      … … … … … ii)

Therefore x + 2y -8 = 0  \text{ and } 2x-y-1= 0 are the two equations.

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Question 6: Find the equations to the sides of an isosceles right angled triangle the equation of whose hypotenuse is 3 x + 4 y = 4 and the opposite vertex is the point (2, 2) .

Answer:

Given equation:

3x+4y=4      \Rightarrow y = \frac{-3}{4} x + 4      \therefore \text{ Slope } m = \frac{-3}{4}

Here x_1  = 2, \hspace{0.5cm} y_1 = 2 , \hspace{0.5cm} \alpha = 45^{\circ}  \hspace{0.5cm} m = \frac{-3}{4}

Therefore the equations are:

y - 2 = \Big(  \frac{ \frac{-3}{4} + \tan 45^{\circ} }{1 - (\frac{-3}{4})  \tan 45^{\circ}} \Big)  ( x - 2)

\Rightarrow y -2 = \Big(  \frac{ \frac{-3}{4} + 1 }{1 + \frac{3}{4} } \Big)  ( x - 2)

\Rightarrow  y -2 = \frac{1}{7} ( x- 2)

\Rightarrow  x-7y+12=0      … … … … … i)

y - 2 = \Big(  \frac{ \frac{-3}{4} - \tan 45^{\circ} }{1 - \frac{3}{4}  \tan 45^{\circ}} \Big)  ( x - 2)

\Rightarrow y -2 = \Big(  \frac{ \frac{-3}{4} - 1 }{1 - \frac{-3}{4} } \Big)  ( x - 2)

\Rightarrow  y -2 = \frac{-7}{1} ( x- 2)

\Rightarrow  7x+y-16=0      … … … … … ii)

Therefore x-7y+12=0  \text{ and } 7x+y-16=0 are the two equations.

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Question 7: The equation of one side of an equilateral triangle is x-y =0 and one vertex is (2+\sqrt{3},5) . Prove that a second side is y + ( 2- \sqrt{3})x =6 and find the equation of the third side.

Answer:

Refer to the adjoining figure. Since the triangle is equilateral triangle, hence all angles are 60^{\circ}

Given equation:

x-y=0      \Rightarrow y = x      \therefore \text{ Slope } m = 1

Here x_1  = 2+\sqrt{3} , \hspace{0.5cm} y_1 = 5 , \hspace{0.5cm} \alpha = 60^{\circ}  \hspace{0.5cm} m = 1

Therefore the equations are:

y - 5 = \Big(  \frac{ 1 + \tan 60^{\circ} }{1 -  \tan 60^{\circ}} \Big)  ( x - 2-\sqrt{3} )

\Rightarrow y -5 = \Big(  \frac{ 1+\sqrt{3} }{1 - \sqrt{3} } \Big)  ( x - 2 - \sqrt{3} )

\Rightarrow  y -5 = -(2 +\sqrt{3})( x - 2 - \sqrt{3} )

\Rightarrow  ( 2+\sqrt{3}) x + y = 5 + ( 2 + \sqrt{3})^2      … … … … … i)

y - 5 = \Big(  \frac{ 1 - \tan 60^{\circ} }{1 +  \tan 60^{\circ}} \Big)  ( x - 2-\sqrt{3} )

\Rightarrow y -5 = \Big(  \frac{ 1-\sqrt{3} }{1 + \sqrt{3} } \Big)  ( x - 2 - \sqrt{3} )

\Rightarrow  y -5 = -(2 -\sqrt{3})( x - 2 - \sqrt{3} )

\Rightarrow  ( 2-\sqrt{3}) x + y -6=0      … … … … … ii)

Therefore ( 2+\sqrt{3}) x + y = 5 + ( 2 + \sqrt{3})^2  \text{ and } ( 2-\sqrt{3}) x + y -6=0 are the two equations.

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Question 8: Find the equations of the two straight lines through (1,2) forming two sides of a square of which 4x + 7y=12 is one diagonal.

Answer:

Refer to the adjoining figure.

Given equation:

4x+7y=12      \Rightarrow y = \frac{-4}{7} x + \frac{12}{7}      \therefore \text{ Slope } m = \frac{-4}{7}

Here x_1  = 1 , \hspace{0.5cm} y_1 = 2 , \hspace{0.5cm} \alpha = 45^{\circ}  \hspace{0.5cm} m = \frac{-4}{7}

Therefore the equations are:

y - 2 = \Big(  \frac{ \frac{-4}{7} + \tan 45^{\circ} }{1 - ( \frac{-4}{7})  \tan 45^{\circ}} \Big)  ( x - 1 )

\Rightarrow y -2 = \Big(  \frac{ \frac{-4}{7}+1 }{1 + \frac{4}{7} } \Big)  ( x - 1 )

\Rightarrow  y -2 = \frac{3}{11}  ( x - 1) 

\Rightarrow  3x-11y+ 19 = 0      … … … … … i)

y - 2 = \Big(  \frac{ \frac{-4}{7} - \tan 45^{\circ} }{1 + ( \frac{-4}{7})  \tan 45^{\circ}} \Big)  ( x - 1 )

\Rightarrow y -2 = \Big(  \frac{ \frac{-4}{7}+1 }{1 + \frac{4}{7} } \Big)  ( x - 1 )

\Rightarrow  y -2 = \frac{-11}{3}  ( x - 1) 

\Rightarrow  11x+3y-17 = 0      … … … … … ii)

Therefore 3x-11y+ 19 = 0  \text{ and } 11x+3y-17 = 0 are the two equations.

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Question 9: Find the equation of two straight line passing through (1,2) and making an angle of 60^{\circ} . with the line x + y = 0 . Find also the area of the triangle formed by the three lines.

Answer:

Refer to the adjoining figure.

Given equation:

x+y=0      \Rightarrow y =  -x      \therefore \text{ Slope } m = -1

Here x_1  = 1 , \hspace{0.5cm} y_1 = 2 , \hspace{0.5cm} \alpha = 60^{\circ}  \hspace{0.5cm} m = -1

Therefore the equations are:

y - 2 = \Big(  \frac{ -1 + \tan 60^{\circ} }{1 +  \tan 60^{\circ}} \Big)  ( x - 1 )

\Rightarrow y -2 = \Big(  \frac{ \sqrt{3}-1 }{\sqrt{3}+1 } \Big)  ( x - 1 )

\Rightarrow  y -2 = (2-\sqrt{3})  ( x - 1)       … … … … … i)

y - 2 = \Big(  \frac{ -1 - \tan 60^{\circ} }{1 -  \tan 60^{\circ}} \Big)  ( x - 1 )

\Rightarrow y -2 = \Big(  \frac{ \sqrt{3}+1 }{\sqrt{3}-1 } \Big)  ( x - 1 )

\Rightarrow  y -2 = (2+\sqrt{3})  ( x - 1)       … … … … … i)

Therefore y -2 = (2-\sqrt{3})  ( x - 1)  \text{ and } y -2 = (2+\sqrt{3})  ( x - 1) are the two equations.

Solving x+y=0 & y -2 = (2-\sqrt{3})  ( x - 1) 

We get x = - \Big( \frac{\sqrt{3}+1}{2} \Big) , y = \Big( \frac{\sqrt{3}+1}{2} \Big)

\therefore B = \Big( - ( \frac{\sqrt{3}+1}{2} \Big) ,  \Big( \frac{\sqrt{3}+1}{2} ) \Big)

Similarly solving x+y = 0 and  y -2 = (2+\sqrt{3})  ( x - 1)

We get x = \Big( \frac{\sqrt{3}-1}{2} \Big) , y = - \Big( \frac{\sqrt{3}-1}{2} \Big)

\therefore C = \Bigg(  \Big( \frac{\sqrt{3}-1}{2} \Big) ,  -\Big( \frac{\sqrt{3}-1}{2} \Big) \Bigg)

AB =\sqrt{  [  1 + ( \frac{\sqrt{3}+1}{2}  )^2   ]  + [ 2 - (\frac{\sqrt{3}+1}{2} )^2   ]  }   = \sqrt{6}

Similarly, BC = AD = \sqrt{6} unit.

Therefore are of \triangle ABC = \frac{\sqrt{3} \times ( \sqrt{6})^2}{4} = \frac{3\sqrt{3}}{2} sq. units.

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Question 10: Two sides of an isosceles triangle are given by the equations 7 x-y+ 3=0 and x +y - 3 = 0 and its third side passes through the point (1, - 10) . Determine the equation of the third side.

Answer:

Refer to the adjoining figure.

AB = BC

\tan B = \tan C

Therefore Slope of AB = 7

and Slope of AC = - 1

Let m be the slope of BC

\therefore \Big| \frac{m-7}{1+7m} \Big | = \Big| \frac{m+1}{1-m} \Big | = \Big| \frac{m+1}{m-1} \Big |

\Rightarrow \frac{m-7}{1+7m} = \pm \frac{m+1}{m-1}

Taking the + sign, we get

m^2 - 8m+7 = 7m^2 + 8m+1

\Rightarrow ( m+3) ( m - \frac{1}{3} ) = 0

\Rightarrow m = - 3, \frac{1}{3}

Also now taking - ve sign,

( m-7)(m-1) = - ( 7m+1)(m+1)

\Rightarrow m^2 - 8m + 7 = - 7m^2 - 8m - 1

\Rightarrow m^2 = - 1 ( not possible)

Therefore the equation of sides are

y + 10 = - 3 ( x-1)

\Rightarrow 3x + y + 7 = 0      … … … … … i)

y + 10 = \frac{1}{3} ( x-1)

\Rightarrow x - 3y - 31 = 0      … … … … … ii)

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Question 11: Show that the point (3, - 5) lies between the parallel lines 2 x+3y-7=0 and 2x+ 3y +12=0 and find the equation of lines through (3,-5) cutting the above lines at an angle of 45^{\circ} .

Answer:

Given lines:

2 x+3y-7=0      … … … … … i)

2x+ 3y +12=0      … … … … … ii)

The distance between line i) and ii)

d = \Big| \frac{12-(-7)}{\sqrt{2^2 + 3^2}} \Big| =  \frac{19}{\sqrt{13}}

Distance for (3, -5) from line i)

d_1 = \Big| \frac{2(3)+3(-5)-7}{\sqrt{2^2 + 3^2}}   \Big| =  \frac{16}{\sqrt{13}}

Distance for (3, -5) from line ii)

d_1 = \Big| \frac{2(3)+3(-5)+12}{\sqrt{2^2 + 3^2}}   \Big| =  \frac{3}{\sqrt{13}}

Now we see that d_1 + d_2 = \frac{19}{\sqrt{13}} = d

Therefore we can say that (3, -5) is between the two given lines.

Let m be the slope of the line passing through ( 3, -5) . This line makes 45^{\circ} with the lines. The slope of the given lines is \frac{-2}{3} . Therefore,

\tan 45^{\circ} = \Big| \frac{m- ( \frac{-2}{3}) }{1 + m ( \frac{-2}{3})} \Big| 

\Rightarrow \frac{3m+2}{3-2m} = \pm 1

Two cases arise: Case 1:  3m + 2 = 3 - 2m  \hspace{0.5cm} \Rightarrow m = \frac{1}{5}

Case 2:  3m + 2 = -(3 - 2m)  \hspace{0.5cm} \Rightarrow m = -5

Therefore, the required lines are

y - ( -5) = \frac{1}{5} (x-3) \hspace{0.5cm} \Rightarrow x - 5y - 28 = 0

and

y - ( -5) = -5 (x-3) \hspace{0.5cm} \Rightarrow 5x + y - 10 = 0 

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Question 12: The equation of the base of an equilateral triangle is x + y =2 and its vertex is (2, - 1) . Find the length and equations of its sides.

Answer:

Since this is an equilateral triangle, hence all the three angles are 60^{\circ} .

The slope of the given lines is -1 . Therefore,

\tan 60^{\circ} = \Big| \frac{m- ( -1) }{1 + m ( -1)} \Big| 

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

Two cases arise: Case 1:  m+1 = \sqrt{3} ( m-1)  \hspace{0.5cm} \Rightarrow m = \frac{\sqrt{3}+1}{\sqrt{3}-1} = 2+\sqrt{3}

Case 2:  m+1 = -\sqrt{3} (m-1)  \hspace{0.5cm} \Rightarrow m = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}

Therefore, the required lines are

y - ( -1) = (2+\sqrt{3}) (x-1) \hspace{0.5cm} \Rightarrow y +1 = (2+\sqrt{3}) (x-2) 

and

y - ( -1) = (2-\sqrt{3}) (x-1) \hspace{0.5cm} \Rightarrow y + 1= (2-\sqrt{3}) (x-2)

Solving x + y =2 and y +1 = (2+\sqrt{3}) (x-2) we get

B = \Big(  \frac{15+\sqrt{3}}{6} , \frac{-(3+\sqrt{3})}{6}   \Big)

Similarly, Solving x + y =2 and y + 1= (2-\sqrt{3}) (x-2) we get

C = \Big(  \frac{15-\sqrt{3}}{6} , \frac{-(3-\sqrt{3})}{6} \Big)

Therefore AB = BC = CA = \sqrt{\frac{2}{3}}

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Question 13: If two opposite vertices of a square arc (1,2) and (5, 8) , find the coordinates of its other two vertices and the equations of its sides.

Answer:

Please refer to the adjoining figure.

Slope of AC = \frac{8-2}{5-1} = \frac{3}{2}

The sides AB and AD passes through (1, 2)   and makes an angle of 45^{\circ}   with AC   whose slope is \frac{3}{2}

The equation of AB and AD are given by

y - 2 =  \frac{\frac{3}{2} \pm \tan 45^{\circ} }{1 \mp \frac{3}{2} \tan 45^{\circ}} ( x - 1)

\Rightarrow  y - 2 = \frac{3 \pm 2}{2 \mp 3} ( x - 1)

\Rightarrow y - 2 = -5( x-1)  \text{ and } y - 2 = \frac{1}{5} ( x-1)

\Rightarrow 5x + y -7 = 0   and x - 5y + 9 = 0

Therefore the equations of AB and AD   are 5x + y -7 = 0   and x - 5y + 9 = 0   respectively.

Since BC is parallel to AD , the equation of BC is x - 5y + \lambda_1 = 0

This line passes through ( 5, 8) . Therefore,

5 - 40 + \lambda_1 = 0 \Rightarrow \lambda_1 = 35

Therefore the equation of BC is x - 5y + 35 = 0

Since CD is parallel to AB , the equation of CD is 5x + y + \lambda_2 = 0

This line passes through ( 5, 8) . Therefore,

25+8 + \lambda_2 = 0 \Rightarrow \lambda_2 = -33

Therefore the equation of CD is 5x + y -33 = 0

Solving equation of AB and BC we get B ( 0,7) .

Solving equation of AD and CD we get D ( 6, 3) .

Hence the two vertices are B ( 0,7) and D ( 6, 3) .