JEE MATHEMATICS MCQ: STRAIGHT LINES

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Application +4 | -1

The equations of sides of a triangle

A B C

are

A B: x+y-1=0, B C: 7 x-y-15=0, A C: x-y-1=0

the equation of angular bisector of internal angle

B

of the triangle is

3 x

+y-k=0

. The positive integer

k

must be.

View answer and explanation

Correct option: A

Option A (5) is correct because solving for the angular bisector of the internal angle B leads to the equation 3x + y - 5 = 0, thus k = 5.

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