JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

x^2-y^2+2 h x y+2 g x+2 f y+c=0

is the locus of points such that it is equidistance from the lines

x+2 y-8=0

and

2 x+y+7=0

, then value of

h+g+f+c

View answer and explanation

Correct option: C

20

Option C (20) is correct because it is found by determining the equation of the angle bisectors and comparing coefficients.

More Questions from STRAIGHT LINES

Practice All Chapter MCQs →

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