JEE MATHEMATICS MCQ: STRAIGHT LINES

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Application +4 | -1

The equation of bisector of that angle between the lines

x+y+1=0

and

2 x-3 y-5=0

which contains the point

(10,-20)

View answer and explanation

Correct option: A

x(\sqrt{13}+2 \sqrt{2})+y (\sqrt{13}-3 \sqrt{2}) +(\sqrt{13}-5 \sqrt{2})=0

Option A (

x(\sqrt{{13}}+2 \sqrt{{2}})+y (\sqrt{{13}}-3 \sqrt{{2}}) +(\sqrt{{13}}-5 \sqrt{{2}})=0

) is correct because it represents the angle bisector that contains the given point (10, -20).

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