Note: When slope (m) and a point (x_1, y_1) is given, the equation of the line is y - y_1 = m ( x-x_1)

Question 1: Find the equation of the straight line passing through the point (6,2) and having slope -3 .

Answer:

Here m = -3   and (x_1, y_1) = ( 6, 2)

Substituting in y - y_1 = m ( x-x_1) we get

y - 2 = - 3( x-6)

\Rightarrow y - 2 = - 3x + 18

\Rightarrow 3x + y - 20 = 0

Hence the equation of the straight line is 3x + y - 20 = 0

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Question 2: Find the equation of the straight line passing through (-2, 3) and inclined at an angle of 45^{\circ} with the x-axis.

Answer:

Here m = \tan 45^{\circ} = 1   and (x_1, y_1) = ( -2, 3)

Substituting in y - y_1 = m ( x-x_1) we get

y - 3 = 1( x+2)

\Rightarrow y-3 = x + 2

\Rightarrow x-y+5=0

Hence the equation of the straight line is x-y+5=0

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Question 3: Find the equation of the line passing through (0, 0) with slope m .

Answer:

Here m = m   and (x_1, y_1) = ( 0,0)

Substituting in y - y_1 = m ( x-x_1) we get

y - 0 = m( x-0)

\Rightarrow y = mx

Hence the equation of the straight line is y=mx

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Question 4: Find the equation of the line passing through (2, 2\sqrt{3}) and inclined with x-axis at an angle of 75^{\circ} .

Answer:

Here \displaystyle \ m = \tan 75^{\circ} =  \tan ( 45^{\circ} + 30^{\circ})

\displaystyle =  \frac{\tan 45^{\circ} + \tan 30^{\circ}}{1 - \tan 45^{\circ} \tan 30^{\circ}}

\displaystyle =  \frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}  \times  \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{3+1 + 2\sqrt{3}}{3-1} = \frac{4+2\sqrt{3}}{2}  = 2 + \sqrt{3}

Also (x_1, y_1) = ( 2, 2\sqrt{3})

Substituting in y - y_1 = m ( x-x_1) we get

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

\Rightarrow y-2\sqrt{3} = (2+\sqrt{3})x - 4 - 2\sqrt{3}

\Rightarrow (2+\sqrt{3})x-y-4=0

Hence the equation of the straight line  is (2+\sqrt{3})x-y-4=0

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Question 5: Find the equation of the straight line which passes through the point (1,2) and makes such an angle with the positive direction of x-axis whose sine is \frac{3}{5}

Answer:

Given \displaystyle \sin \theta = \frac{3}{5}

We know,

\displaystyle \tan \theta=  \frac{\sin \theta}{1 - \sin^2 \theta }  =  \frac{3/5}{\sqrt{1 - 9/25}}  =  \frac{3}{\sqrt{16}} = \frac{3}{4}

Therefore \displaystyle m =  \frac{3}{4}   and (x_1, y_1) = ( 1, 2)

Substituting in y - y_1 = m ( x-x_1) we get

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

\Rightarrow 3x-4y+5=0

Hence the equation of the straight line is 3x-4y+5=0

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Question 6: Find the equation of the straight line passing through (3, -2) and making an angle of 60^{\circ} with the positive direction of y-axis

Answer:

\displaystyle \text{ Slope of line  } = m = \tan 30^{\circ} =  \frac{1}{\sqrt{3}}    and  (x_1, y_1) = ( 3, -2)

Substituting in y - y_1 = m ( x-x_1) we get

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

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

\Rightarrow x-\sqrt{3} y - 3 - 2\sqrt{3} = 0

Hence the equation of the straight line is x-\sqrt{3} y - 3 - 2\sqrt{3}=0

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Question 7: Find the lines through the point (0,2) making angle \frac{\pi}{3} and \frac{2\pi}{3} with the x-axis. Also, find the lines parallel to them cutting the y-axis at a distance of 2 units below the origin.

Answer:

Given \displaystyle m_1 = \tan  \frac{\pi}{3}  = \sqrt{3}

\displaystyle m_2 = \tan  \frac{2\pi}{3}  = - \tan  \frac{\pi}{3}  = -\sqrt{3}

Also (x_1, y_2) = ( 0, 2)

Therefore the equation passing through ( 0,2) and having m_1 and m_2 slope respectively are

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

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

Now let us find lines parallel to these lines but passing through ( 0, -2)

Therefore the equation passing through ( 0,-2) and having m_1 and m_2 slope respectively are

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

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

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Question 8: Find the equations of the straight lines which cut off an intercept 5 from the y-axis and are equally inclined to the axes.

Answer:

Given the lines are equally inclined to the axes.

Hence, their inclination with positive x-axis are 45^{\circ} and 135^{\circ}

Therefore for Line 1: m_1 = \tan 45^{\circ} = 1

and for Line 2: m_2 = \tan 135^{\circ} = - \tan 45^{\circ} = -1

Hence the equation of Line 1:  y - 5 = 1 ( x - 0) \Rightarrow x - y + 5 = 0

And the equation for Line 2: y - 5 = -1 ( x - 0) \Rightarrow x + y -5 = 0

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Question 9: Find the equation of the line which intercepts a length 2 on the positive direction of the x-axis and is inclined at an angle of 135^{\circ} with the positive direction of y-axis.

Answer:

Here m = \tan 45^{\circ} = 1 and (x_1, y_1) = ( 2, 0)

Therefore equation of the line is: y-0 = 1 ( x-2) \Rightarrow x-y - 2 = 0

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Question 10: Find the equation of the straight line which divides the join of the points (2, 3) and (- 5, 8) in the ratio 3 : 4 and is also perpendicular to it.

Answer:

Given points A(2, 3) and B(- 5, 8)

\displaystyle \text{ Therefore slope of  } AB =  \frac{8-3}{-5-2}  =  \frac{-5}{7}

\displaystyle \text{ Therefore Slope of line perpendicular to  } AB =  \frac{-1}{(-5/7)}  =  \frac{7}{5}

The coordinate of the point that divides AB in the ratio of 3:4 is

\displaystyle (x_1, y_1) = \Big(  \frac{3(-5) + 4(2)}{3+4}  ,  \frac{3(8)+4(3)}{3+4}  \Big) = \Big(  \frac{-7}{7} , \frac{36}{7}  \Big) =\Big( -1,  \frac{36}{9}  \Big)

Therefore the equation of the required line:

\displaystyle y -  \frac{36}{7}  =  \frac{7}{5}  ( x+1)

\Rightarrow 7y - 36 = \frac{7}{5} ( 7x + 7)

\Rightarrow 35y - 180 = 49x + 49

\Rightarrow 49x - 35y + 229 = 0

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Question 11: Prove that the perpendicular drawn from the point (4, 1) on the join of (2, -1) and (6,5) divides it in the ratio 5 : 8 .

Answer:

Given points A(2, -1) and B(6,5 )

\displaystyle \text{ Slope of  } AB =  \frac{5-(-1)}{6-2} = \frac{6}{4} = \frac{3}{2}

PQ is \perp to AB . Therefore slope of \displaystyle PQ =  \frac{-1}{3/2} = \frac{-2}{3}

Therefore equation of PQ is:2021-01-07_15-29-43

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

\Rightarrow 3y - 3 = - 2x + 8

\Rightarrow 2x + 3y = 11      … … … … … i)

Equation of AB :

\displaystyle y - 5 = \frac{3}{7}  (x-6)

\Rightarrow 2y - 10 = 3x - 18

\Rightarrow 3x - 2y = 8   … … … … … ii)

Point of intersection will be found by solving the two equations.

Multiply i) by 3 and ii) by 2 and then subtract ii) from i) we get

{ \hspace{1.0cm} 3 \times (2x + 3y = 11)} \\ \underline {(-1) \ 2 \times ( 3x - 2y = 8)} \\ {\hspace{2.5cm}13y = 17}

\displaystyle \Rightarrow y =  \frac{17}{13}

Substituting in i) we get

\displaystyle 2x + 3 \Big(  \frac{17}{13}  \Big ) = 11

\displaystyle \Rightarrow x =  \frac{46}{13}

Hence the point of intersection is

\displaystyle \Big( \frac{46}{13} ,  \frac{17}{13} \Big) = \Big( \frac{5(6) + 8 ( 2) }{5+8} , \frac{5(5) + 8 ( -1) }{5+8} \Big)

Therefore we see that  Q \Big( \frac{46}{13} ,  \frac{17}{13} \Big) divides AB in the ratio of 5:8

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Question 12: Find the equations to the altitudes of the triangle whose angular points ate A(2,-2), B (1, 1) and C (- 1, 0) .

Answer:

2021-01-07_15-23-04Given A(2,-2), B (1, 1) and C (- 1, 0) .

Let AD , BE and CF are the altitudes as shown in the figure.

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

\displaystyle \text{ Therefore slope of } AD =   \frac{-1}{(1/2)}   = -2

Therefore equation of AD :

y - ( -2) = -2 ( x-2)

\Rightarrow y + 2 = - 2x + 4

\Rightarrow 2x + y + 2 = 0

\displaystyle \text{ Slope of  } AC =  \frac{0-(-2)}{-1-2} = \frac{2}{-3} = \frac{-2}{3}

\displaystyle \text{ Therefore slope of  } BE =   \frac{-1}{(-2/3)} = \frac{3}{2} 

Therefore equation of BE :

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

\Rightarrow 2y - 2 = 3x - 3

\Rightarrow 3x-2y-1=0

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

\displaystyle \text{ Therefore slope of  } CF =   \frac{-1}{-3}   =   \frac{1}{3} 

Therefore equation of CF :

\displaystyle y - 0 =  \frac{1}{2}  ( x+1)

\Rightarrow 3y = x + 1

\Rightarrow x-3y+1=0

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Question 13: Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1,2) .

Answer:

Given points A(3, 4) and B ( -1, 2)

Therefore the midpoint of \displaystyle AB = (  \frac{-1+3}{2}  ,  \frac{2+4}{2}  ) = ( 1, 3)

Slope of \displaystyle AB =  \frac{2-4}{-1-3} = \frac{-2}{-4}  \frac{1}{2}

Therefore the slope of the bisector \displaystyle - \frac{-1}{1/2}  = -2

Hence the equation of the bisector:

y-3 = -2( x-1)

\Rightarrow y - 3 = -2x + 2

\Rightarrow 2x + y - 5 = 0

Hence the equation of the right bisector is 2x + y - 5 = 0

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Question 14: Find the equation of the line passing through the point (- 3, 5) and perpendicular to the line joining (2,5) and (-3, 6) .

Answer:

Given points A( 2, 5) and B ( -3, 6)

Slope of \displaystyle AB =  \frac{6-5}{-3-2}  =  \frac{1}{-5} = \frac{=1}{5}

Therefore the slope of the perpendicular line \displaystyle =  \frac{-1}{(-1/5)}  = 5

Hence the equation of the perpendicular which is passing through ( -3, 5) :

y - 5 = 5 ( x+3)

\Rightarrow 5x - y + 20 = 0

Hence the equation of the line is 5x - y + 20 = 0

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Question 15: Find the equation of the right bisector of the line segment joining the points A(1, 0) and B (2, 3) .

Answer:

Given points A(1,0) and B( 2, 3)

Mid point of \displaystyle AB = \Big(  \frac{2+1}{2} , \frac{3+0}{2}   \Big) = \Big(   \frac{3}{2}   ,   \frac{3}{2}   \Big)

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

Therefore slope of bisector \displaystyle =   \frac{-1}{3} 

Therefore the equation of the right bisector:

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

\displaystyle \Rightarrow 6y - 9 = - 2 \Big( x -   \frac{3}{2}   \Big)

\Rightarrow 6y - 9 = - 2x + 3

\Rightarrow 2x + 6y - 12 = 0

\Rightarrow x + 3y - 6 = 0

Hence the equation of the right bisector is  x + 3y - 6 = 0