Question 1: Reduce the equation \sqrt{3}x +y +2 = 0   to:

(i) slope-intercept form and find slope and y-intercept     (ii) intercept form and find intercept on the axes;     (iii) the normal form and find p and \alpha .

Answer:

i)       Given \sqrt{3}x +y +2 = 0

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

Therefore Slope = -\sqrt{3} and y-intercept = -2

ii)      Given \sqrt{3}x +y +2 = 0

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

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

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

Therefore x-intercept (a)  = \frac{-2}{\sqrt{3}} and y-intercept (b) = -2

iii)     Given \sqrt{3}x +y +2 = 0

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

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

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

Therefore p = 1, \hspace{0.5cm} \cos \alpha = \frac{-\sqrt{3}}{2} , \hspace{0.5cm} \sin \alpha = \frac{-1}{2}    and   \alpha = 210^{\circ}

Since the coefficient of both x and y are negative, \alpha lies in the 3^{rd} Quadrant.

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Question 2: Reduce the following equations to the normal form and find p and \alpha in each case:

i) x+\sqrt{3}y - 4 = 0      ii) x+y + \sqrt{2} = 0      iii) x-y + 2\sqrt{2} = 0

iv) x-3=0      v) y-2=0

Answer:

i)       Given x+\sqrt{3}y - 4 = 0

\Rightarrow x+\sqrt{3}y  = 4 

\Rightarrow \frac{x}{\sqrt{(1)^2 + ( \sqrt{3} )^2}} + \frac{\sqrt{3} y}{\sqrt{(1)^2 + ( \sqrt{3} )^2}}  =  \frac{4}{\sqrt{(1)^2 + ( \sqrt{3} )^2}}

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

Therefore p = 2, \hspace{0.5cm} \cos \alpha = \frac{1}{2} , \hspace{0.5cm} \sin \alpha = \frac{\sqrt{3}}{2}    and   \alpha = 60^{\circ}

Since the coefficient of both x and y are positive, \alpha lies in the 1^{st} Quadrant.

ii)     Given x+y + \sqrt{2} = 0

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

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

\Rightarrow - \frac{x}{\sqrt{2} } - \frac{y}{\sqrt{2}} = 1

Therefore p = 1, \hspace{0.5cm} \cos \alpha = \frac{-1}{\sqrt{2}} , \hspace{0.5cm} \sin \alpha = \frac{-1}{\sqrt{2}}    and   \alpha = 225^{\circ}

Since the coefficient of both x and y are negative, \alpha lies in the 3^{rd} Quadrant.

iii)    Given x-y + 2\sqrt{2} = 0

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

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

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

Therefore p = 2, \hspace{0.5cm} \cos \alpha = \frac{-1}{2} , \hspace{0.5cm} \sin \alpha = \frac{1}{\sqrt{2}}    and   \alpha = 135^{\circ}

Since the coefficient of both x is negative and y y is positive, \alpha lies in the 2^{nd} Quadrant.

iv)    Given x-3=0

\Rightarrow x + 0y = 3

\Rightarrow \frac{x}{\sqrt{(1)^2 + ( 0 )^2}} + \frac{ 0 \cdot y}{\sqrt{(1)^2 + ( 0 )^2}}  =  \frac{3}{\sqrt{(1)^2 + ( 0 )^2}}

\Rightarrow \frac{x}{1} + \frac{0 \cdot y}{1} = 3

Therefore p = 3, \hspace{0.5cm} \cos \alpha = 1   , \hspace{0.5cm} \sin \alpha = 0     and   \alpha = 0^{\circ}

v)      Given y-2=0

\Rightarrow 0x + 1y = 2

\Rightarrow \frac{0 \cdot x}{\sqrt{(0)^2 + ( 1 )^2}} + \frac{ y}{\sqrt{(0)^2 + ( 1 )^2}}  =  \frac{2}{\sqrt{(0)^2 + ( 1 )^2}}

\Rightarrow \frac{0 \cdot x}{1} + \frac{y}{1} = 2

Therefore p = 2, \hspace{0.5cm} \cos \alpha = 0   , \hspace{0.5cm} \sin \alpha = 1     and   \alpha = 90^{\circ}

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Question 3: Put the equation  \frac{x}{a} + \frac{y}{b} =1 to the slope intercept form and find its slope and y-intercept.

Answer:

Given \frac{x}{a} + \frac{y}{b} =1

\Rightarrow bx + ay = ab

\Rightarrow ay = - bx + ab

\Rightarrow y = \frac{-b}{a}   x + b

Comparing with y = mx + c where slope = m and y-intercept = c

Therefore \text{Slope} = \frac{-b}{a} and y-intercept = b

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Question 4: Reduce the lines 3 x - 4 y + 4 =0 and 2 x + 4 y -5 = 0 to the normal form and hence find which line is nearer to the origin.

Answer:

Line 1: Given 3x-4y+4=0

\Rightarrow -3x+4y=4

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

\Rightarrow - \frac{3x}{5 } + \frac{4y}{5} = \frac{4}{5}

Therefore p_1 = \frac{4}{5}

Line 2: Given 2x+4y-5=0

\Rightarrow 2x+4y=5

\Rightarrow \frac{2x}{\sqrt{(2)^2 + ( 4 )^2}} + \frac{ 4y}{\sqrt{(2)^2 + ( 4 )^2}} = \frac{5}{\sqrt{(-3)^2 + ( 4 )^2}}

\Rightarrow  \frac{2x}{2\sqrt{5} } + \frac{2y}{2\sqrt{5}} = \frac{\sqrt{5}}{2}

Therefore p_2 = \frac{\sqrt{5}}{2}

Since \frac{\sqrt{5}}{2} > \frac{4}{5} , we can say that 3x-4y+4 = 0 is nearer to the origin as compared to 2x+ 4y -5=0

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Question 5: Show that the origin is equidistant from the lines 4x + 3y + 10 = 0 ; 5x -12y + 26 =0 and 7x+24y=50 .

Answer:

Line 1: Given 4x + 3y + 10 = 0

\Rightarrow -4x-3y=10

\Rightarrow \frac{-4x}{\sqrt{(-4)^2 + ( -3 )^2}} - \frac{ 3y}{\sqrt{(-4)^2 + ( -3 )^2}} = \frac{10}{\sqrt{(-4)^2 + ( -3 )^2}}

\Rightarrow - \frac{4x}{5 } - \frac{3y}{5} = 2

Therefore p_1 = 2  

Line 2: Given 5x -12y + 26 =0

\Rightarrow -5x+12y=26

\Rightarrow \frac{-5x}{\sqrt{(-5)^2 + ( 12 )^2}} + \frac{ 12y}{\sqrt{(-5)^2 + ( 12 )^2}} = \frac{26}{\sqrt{(-5)^2 + ( 12 )^2}}

\Rightarrow  - \frac{5x}{13 } + \frac{12y}{13} = 2  

Therefore p_2 = 2  

Line 3: Given 7x+24y=50

\Rightarrow \frac{7x}{\sqrt{(7)^2 + ( 24 )^2}} + \frac{ 24y}{\sqrt{(7)^2 + ( 24 )^2}} = \frac{50}{\sqrt{(7)^2 + ( 24 )^2}}

\Rightarrow  \frac{7x}{25 } + \frac{24y}{25} = 2  

Therefore p_3 = 2  

Since p_1=p_2=p_3 , the given lines 4x + 3y + 10 = 0 ; 5x -12y + 26 =0 and 7x+24y=50 are equidistance from the origin.

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Question 6: Find the values of 0 and p , if the equation x \cos \theta + y \sin \theta =p is the normal form of the line \sqrt{3}x+y+2=0 .

Answer:

Given \sqrt{3}x +y +2 = 0

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

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

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

Therefore p = 1, \hspace{0.5cm} \cos \theta = \frac{-\sqrt{3}}{2} , \hspace{0.5cm} \sin \theta = \frac{-1}{2}    and   \theta = 210^{\circ}

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Question 7: Reduce the equation 3 x -2y + 5 =0 to the intercept form and find the x and y intercepts.

Answer:

Given equation 3 x -2y + 6 =0

\Rightarrow 3x-2y=-6

\Rightarrow \frac{3x}{-6} - \frac{2y}{-6} = 1

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

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

Hence, x-intercept \Rightarrow = -2 and y-intercept \Rightarrow = 3

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Question 8: The perpendicular distance of a line from the origin is 5 units and its slope is - 1 . Find the equation of the line.

Answer:

Given p = 5

Therefore equation of line:

x \cos \alpha + y \sin \alpha = 5

\Rightarrow y \sin \alpha = - x \cos \alpha + 5

\Rightarrow y = - x \cot \alpha + \frac{5}{\sin \alpha}

Comparing with y = mx + c where slope = m and y-intercept = c

Slope = - \cot \alpha  = - 1

\Rightarrow \cot \alpha = 1 \Rightarrow \alpha = 45^{\circ}

Therefore the equation of line is

x \cos 45^{\circ} + \sin 45^{\circ} = 5

\Rightarrow \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 5

\Rightarrow x + y = 5\sqrt{2}