The distance between the parallel lines $3 x-4 y+7=0$ and $3 {x}-4 {y}+5=0$ is $\frac{{a}}{{b}}$. Value of ${a}+{b}$ is

JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

The distance between the parallel lines

3 x-4 y+7=0

and

3 {x}-4 {y}+5=0

\frac{{a}}{{b}}

. Value of

{a}+{b}

View answer and explanation

Correct option: A

Option A (7) is correct because the distance between parallel lines is calculated using the formula, and the sum of the numerator and denominator of the simplified distance is 7.

More Questions from STRAIGHT LINES

Practice All Chapter MCQs →

If the three lines

x-3 y=p, a x+2 y=q

and

a x+y=r

form a right-angled triangle then :
The orthocentre of the triangle whose vertices are

\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\}

is
The distance between the parallel lines

y = 2\;x + 4

and

6x = 3y + 5

is The distance between the pair of lines given by

{x^2} + {y^2} + 2xy - 8ax - 8ay - 9{a^2} = 0

is Find the image of the point

(3,8)

with respect of the line

x+3 y=7

assuming the line to be plane mirror: The centroid of the triangle whose three sides are given by the combined equation

\left( {{x^2} + 7xy + 2{y^2}} \right)\left( {y - 1} \right) = 0