The normal form of the equation $x-\sqrt{3} y+8=0$ is

$x \cos 120^{\circ}+y \sin 120^{\circ}=8$

JEE MATHEMATICS MCQ: STRAIGHT LINES

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The normal form of the equation

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

View answer and explanation

Correct option: B

x \cos 120^{\circ}+y \sin 120^{\circ}=4

Option B (

x \cos 120^{\circ}+y \sin 120^{\circ}=4

) is correct because it represents the given equation in normal form, where the coefficients of x and y are cosine and sine of the angle, respectively, and the right-hand side is the perpendicular distance from the origin.

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is
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y = 2\;x + 4

and

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is The distance between the pair of lines given by

{x^2} + {y^2} + 2xy - 8ax - 8ay - 9{a^2} = 0

is Find the image of the point

(3,8)

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x+3 y=7

assuming the line to be plane mirror: The centroid of the triangle whose three sides are given by the combined equation

\left( {{x^2} + 7xy + 2{y^2}} \right)\left( {y - 1} \right) = 0