JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

Assertion: The distance between the parallel lines

3 x-4 y+9=0 \text { and } 6 x-8 y-15=0 \text { is } \frac{33}{10}

Reason: Distance between the parallel lines

{Ax}+{By}+{C}_1=0

and

{Ax}+{By}+{C}_2=0

, is given by

{d}=\frac{\left|{C}_1-{C}_2\right|}{\sqrt{{A}^2+{B}^2}}

View answer and explanation

Correct option: A

Assertion is correct, reason is correct; reason is a correct
explanation for assertion

Option A is correct because the assertion is true, the reason is true, and the reason correctly explains the assertion, as shown by calculating the distance between the parallel lines.

More Questions from STRAIGHT LINES

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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