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The equation of a line whose perpendicular distance from the origin is 8 units and the angle made by perpendicular with positive x-axis is 60 degree is:
- a)5x - 3y + 6 = 0
- b)x + √3y = 16
- c)√3x + y - 2 = 0
- d)√3x + √2y - 5 = 0
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The equation of a line whose perpendicular distance from the origin is...
If p isthe length of the normal from the origin to a line and ωis the angle made by the normal with the positive direction of thex-axis,then the equation of the line is given by xcosω +ysinω= p.
Here, p = 8 units and ω= 60°
Thus, therequired equation of the given line is
xcos 60° + y sin 60° = 8
x(1/2) + y(√3/2) = 8
x/2 + √3y/2 = 8
x + √3y = 16
Here, p = 8 units and ω= 60°
Thus, therequired equation of the given line is
xcos 60° + y sin 60° = 8
x(1/2) + y(√3/2) = 8
x/2 + √3y/2 = 8
x + √3y = 16
Most Upvoted Answer
The equation of a line whose perpendicular distance from the origin is...
If p isthe length of the normal from the origin to a line and ωis the angle made by the normal with the positive direction of thex-axis,then the equation of the line is given by xcosω +ysinω= p.
Here, p = 8 units and ω= 60°
Thus, therequired equation of the given line is
xcos 60° + y sin 60° = 8
x(1/2) + y(√3/2) = 8
x/2 + √3y/2 = 8
x + √3y = 16
Here, p = 8 units and ω= 60°
Thus, therequired equation of the given line is
xcos 60° + y sin 60° = 8
x(1/2) + y(√3/2) = 8
x/2 + √3y/2 = 8
x + √3y = 16
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Community Answer
The equation of a line whose perpendicular distance from the origin is...
Here you can use polar form of line ...i.e xcos(¢)+ysin(¢)=phere ¢=60 p=8 xcos60+ysin60=8 x/2+√3y/2=8x+√3y=16...that is Ans...
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The equation of a line whose perpendicular distance from the origin is 8 units and the angle made by perpendicular with positive x-axis is 60 degree is:a)5x - 3y + 6 = 0b)x +√3y = 16c)√3x + y -2 = 0d)√3x + √2y - 5= 0Correct answer is option 'B'. Can you explain this answer? for JEE 2026 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of a line whose perpendicular distance from the origin is 8 units and the angle made by perpendicular with positive x-axis is 60 degree is:a)5x - 3y + 6 = 0b)x +√3y = 16c)√3x + y -2 = 0d)√3x + √2y - 5= 0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of a line whose perpendicular distance from the origin is 8 units and the angle made by perpendicular with positive x-axis is 60 degree is:a)5x - 3y + 6 = 0b)x +√3y = 16c)√3x + y -2 = 0d)√3x + √2y - 5= 0Correct answer is option 'B'. Can you explain this answer?.
The equation of a line whose perpendicular distance from the origin is 8 units and the angle made by perpendicular with positive x-axis is 60 degree is:a)5x - 3y + 6 = 0b)x +√3y = 16c)√3x + y -2 = 0d)√3x + √2y - 5= 0Correct answer is option 'B'. Can you explain this answer? for JEE 2026 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of a line whose perpendicular distance from the origin is 8 units and the angle made by perpendicular with positive x-axis is 60 degree is:a)5x - 3y + 6 = 0b)x +√3y = 16c)√3x + y -2 = 0d)√3x + √2y - 5= 0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of a line whose perpendicular distance from the origin is 8 units and the angle made by perpendicular with positive x-axis is 60 degree is:a)5x - 3y + 6 = 0b)x +√3y = 16c)√3x + y -2 = 0d)√3x + √2y - 5= 0Correct answer is option 'B'. Can you explain this answer?.
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