JEE MATHEMATICS MCQ: STRAIGHT LINES

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Analyzing +4 | -1

A line through

A(-5,-4)

meets the lines

x+3 y+2=0,2 x+y+4=0

and

x-y-5=0

B, C

and

D

respectively. If

\left(\frac{15}{A B}\right)^2+\left(\frac{10}{A C}\right)^2=\left(\frac{6}{A D}\right)^2

, then the equation of the line is

View answer and explanation

Correct option: A

2 x+3 y+22=0

Option A (

2 x+3 y+22=0

) is correct because it satisfies the given condition involving distances AB, AC, and AD.

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