JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

The equation of the bisector of that angle between the lines

x+y=3

and

2 x-y=2

which contains the point

(1,1)

View answer and explanation

Correct option: A

(\sqrt{5}-2 \sqrt{2}) x+(\sqrt{5}+\sqrt{2}) y-3 \sqrt{5}+2 \sqrt{2}=0

Option A (

\sqrt{5}-2 \sqrt{2}) x+(\sqrt{5}+\sqrt{2}) y-3 \sqrt{5}+2 \sqrt{2}=0

) is correct because it represents the angle bisector that contains the point (1,1) after proper normalization.

More Questions from STRAIGHT LINES

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

a x+y=r

\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