JEE MATHEMATICS MCQ: STRAIGHT LINES

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Application +4 | -1

Let P=(-1,0),Q=(0,0) and

R = \left( {3,\;3\sqrt 3 } \right)

be three points. The equation of the bisector of the angle PQR is

View answer and explanation

Correct option: A

\sqrt 3 x + y = 0

Option A (

\sqrt 3 x + y = 0

) is correct because the bisector's equation is derived from the angle between the lines and the slope.

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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

(3,8)

x+3 y=7

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