JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

The equation of the line passing through the intersection of

x - \sqrt 3 \;y + \sqrt 3 - 1 = 0

and

x + y - 2 = 0

and making an angle of

{15^ \circ }

with the first line is

View answer and explanation

Correct option: A

x - y = 0

Option A (

x - y = 0

) is correct because it represents a line passing through the intersection of the given lines and making a 15-degree angle with the first line.

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x-3 y=p, a x+2 y=q

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\left\{ {a{t_1}{t_2},\;a\left( {{t_1} + {t_2})\} ,\{ a{t_2}{t_3},a({t_2} + {t_3}} \right)} \right\},\;\left\{ {a{t_3}{t_1},\;a\left( {{t_3} + {t_1}} \right)} \right\}

y = 2\;x + 4

6x = 3y + 5

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

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x+3 y=7

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