JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

The angle which the normal to the line

x-\sqrt{3} y+8

=0

passing through the origin, makes with the positive

x

-axis is

View answer and explanation

Correct option: C

120^{\circ}

Option C (

120^{\circ}

) is correct because it is the angle the normal to the line makes with the positive x-axis, derived from the normal form of the line equation.

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6x = 3y + 5

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

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