JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Medium

1 of 1

Application +4 | -1

A point equidistant from the lines

4\;x + 3\;y + 10 = 0,\;5\;x - 12\;y + 26 = 0

and

7\;x + 24\;y - 50 = 0

View answer and explanation

Correct option: C

(0,0)

The correct answer is C (0,0) because the perpendicular distance from (0,0) to each of the given lines is the same.

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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