JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

The point on the axis of

x

, whose perpendicular distance from the straight line

\frac{x}{a} + \frac{y}{b} = 1

a

, are

View answer and explanation

Correct option: B

\left( {\frac{a}{b}\left( {\;b \pm \sqrt {{a^2} + {b^2}} } \right),\;0} \right)

The correct answer is B ((

\frac{a}{b}

(b ± √(a²+b²)), 0)) because the perpendicular distance from a point on the x-axis to the given line is calculated and set equal to 'a'.

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