straight-line-in-normal-formNote: The equation of the line whose length of the perpendicular from the origin is p and the angle made by the perpendicular with the positive x-axis is given by \alpha is given by x \cos \alpha + y \sin \alpha = p

Question 1: Find the equation of a line for which

i) p = 5, \alpha = 60^{\circ}      ii) p = 4, \alpha = 150^{\circ}      iii) p = 8, \alpha = 225^{\circ}      vi) p = 8, \alpha = 300^{\circ}

Answer:

i)       Given p = 5, \alpha = 60^{\circ}

Therefore the equation of line in normal form:

\therefore x \cos 60^{\circ} + y \sin 60^{\circ} = 5

\Rightarrow \frac{x}{2}  + \frac{\sqrt{3}y}{2}  = 5

\Rightarrow x + \sqrt{3} y = 10

ii)      Given p = 4, \alpha = 150^{\circ}

Therefore the equation of line in normal form:

\therefore x \cos 150^{\circ} + y \sin 150^{\circ} = 4

\Rightarrow x \cos (180^{\circ}-30^{\circ}) + y \sin (180^{\circ}-30^{\circ}) = 4

\Rightarrow -x \cos 30^{\circ} + y \sin 30^{\circ} = 4

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

\Rightarrow \sqrt{3}x - y +8=0

iii)    Given p = 8, \alpha = 225^{\circ}

Therefore the equation of line in normal form:

\therefore x \cos 225^{\circ} + y \sin 225^{\circ} = 8

\Rightarrow x \cos (180^{\circ}+45^{\circ}) + y \sin (180^{\circ}+45^{\circ}) = 8

\Rightarrow -x \cos 45^{\circ} - y \sin 45^{\circ} = 8

\Rightarrow \frac{- x}{\sqrt{2}} - \frac{y}{\sqrt{2}} = 8

\Rightarrow x+y + 8\sqrt{2} = 0

vi)     Given p = 8, \alpha = 300^{\circ}

Therefore the equation of line in normal form:

\therefore x \cos 300^{\circ} + y \sin 200^{\circ} = 8

\Rightarrow x \cos (360^{\circ}-60^{\circ}) + y \sin (360^{\circ}-60^{\circ}) = 8

\Rightarrow x \cos 60^{\circ} - y \sin 60^{\circ} = 8

\Rightarrow \frac{x}{2} - \frac{\sqrt{3}y}{2} = 8

\Rightarrow x-\sqrt{3} y = 16

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Question 2: Find the equation of the line on which the length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of x-axis is 30^{\circ} .

Answer:

Given p = 4, \alpha = 30^{\circ}

Therefore the equation of line in normal form:

\therefore x \cos 30^{\circ} + y \sin 30^{\circ} = 4

\Rightarrow \frac{\sqrt{3} x}{2} + \frac{y}{2} = 4

\Rightarrow \sqrt{3}x + y = 8

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Question 3: Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with the positive direction of x-axis is 15^{\circ} .

Answer:

Given p = 4, \alpha = 15^{\circ}

Therefore the equation of line in normal form:

\therefore x \cos 15^{\circ} + y \sin 15^{\circ} = 4

\therefore x \cos ( 45^{\circ} - 30^{\circ}) + y \sin ( 45^{\circ} - 30^{\circ}) = 4

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

\Rightarrow (\sqrt{3}+1) x + (\sqrt{3}-1)y = 8\sqrt{2}

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Question 4: Find the equation of the straight line at a distance of 3 units from the origin such  that the perpendicular from the origin to the line makes an angle \alpha given by \tan \alpha = \frac{5}{12} with the positive direction of x-axis.

Answer:

Given p = 3, \alpha = \tan^{-1} \Big( \frac{5}{12} \Big)

\therefore \tan \alpha = \frac{5}{12}       \Rightarrow \frac{\sin \alpha}{\cos \alpha} = \frac{5}{12}       \Rightarrow \frac{\sin \alpha}{\cos \alpha} = \frac{5/13}{12/13}

\therefore  \sin \alpha = \frac{5}{13}   and  \cos \alpha = \frac{12}{13}

Therefore the equation of line in normal form:

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

\Rightarrow x \Big( \frac{12}{13} \Big) + y \Big( \frac{5}{13} \Big) = 3

\Rightarrow 12x + 5y = 39

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Question 5: Find the equation of the straight line on which the length of the perpendicular from the origin is 2 and the perpendicular makes an angle \alpha with x-axis such that \sin \alpha = \frac{1}{3} .

Answer:

Given p = 2, \sin \alpha = \frac{1}{3}

\therefore \cos \alpha = \sqrt{ 1 - \sin^2 \alpha } = \sqrt{1 - \frac{1}{9}} = \frac{2\sqrt{2}}{3}

Therefore the equation of line in normal form:

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

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

\Rightarrow 2\sqrt{2}x + y = 6

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Question 6: Find the equation of the straight line upon which the length of the perpendicular from the
origin is 2 and the slope of this perpendicular is \frac{5}{12} .

Answer:

Given p = 2, \tan \alpha = \frac{5}{12}

\therefore \sin \alpha = \frac{5}{13}   and  \cos \alpha = \frac{12}{13}

Therefore the equations of line in normal form:

Case 1: \alpha = \alpha

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

\Rightarrow x \Big( \frac{12}{13} \Big) + y \Big( \frac{5}{13} \Big) = 2

\Rightarrow 12x + 5y = 26

Case 2: \alpha = (180^{\circ}+ \alpha)

x \cos (180^{\circ}+ \alpha) + y \sin (180^{\circ}+ \alpha) = p

- x \cos \alpha - y \sin \alpha = 2

\Rightarrow -x \Big( \frac{12}{13} \Big) - y \Big( \frac{5}{13} \Big) = 2

\Rightarrow -12x - 5y = 26

\Rightarrow 12x  + 5y - 26=0

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Question 7: The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 15^{\circ} with the positive direction of y-axis. Find the equation of the line.

Answer:

2021-01-12_23-44-55Given p = 7, \alpha = 30^{\circ}

Therefore the equation of line in normal form:

\therefore x \cos 30^{\circ} + y \sin 30^{\circ} = 7

\Rightarrow \frac{\sqrt{3} x}{2} + \frac{y}{2} = 7

\Rightarrow \sqrt{3} x + y = 14

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Question 8: Find the value of \theta   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 equation is

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

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

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

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

Comparing this with the equation x \cos \theta + y \sin \theta  = p we get

p = 1,

\cos \theta = \frac{-\sqrt{3}}{2}       \Rightarrow \theta = 210^{\circ}

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Question 9: Find the equation of the straight line which makes a triangle of area 96\sqrt{3} with the axes and perpendicular from the origin to it makes an angle of 30^{\circ} with y-axis.

Answer:

2021-01-12_23-38-52Given p = 7, \alpha = 60^{\circ}

Therefore the equation of line in normal form:

\therefore x \cos 60^{\circ} + y \sin 60^{\circ} = p

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

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

Therefore x-intercept = 2p    and   y-intercept = \frac{2p}{\sqrt{3}}

It is given that the area of \triangle AOB = 96\sqrt{3}

\therefore \frac{1}{2} \times 2p \times \frac{2p}{\sqrt{3}} = 96\sqrt{3}

\Rightarrow p^2 = 144

\Rightarrow p = 12

Therefore equation of AB = x + \sqrt{3}y = 24

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Question 10: Find the equation of a straight line on which the perpendicular from the origin makes an angle of 30^{\circ} with x-axis and which forms a triangle of area 50\sqrt{3} with the axes

Answer:

2021-01-12_23-38-21Given OL = p, \alpha = 30^{\circ}

Therefore the equation of line in normal form:

\therefore x \cos 30^{\circ} + y \sin 30^{\circ} = p

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

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

Therefore x-intercept = \frac{2p}{\sqrt{3}}   and   y-intercept = 2p

It is given that the area of \triangle AOB = \frac{50}{\sqrt{3}}

\therefore \frac{1}{2} \times \frac{2p}{\sqrt{3}} \times 2p = \frac{50}{\sqrt{3}}

\Rightarrow p^2 = 25

\Rightarrow p = 5

Therefore equation of AB = \sqrt{3} x + y = 10