(A) The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula $d = \frac{|C_1 - C_2|}{\sqrt{A^2

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(A) The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.Here,$A = 3$,$B = -4$,$C_1 = 7$,and $C_2 = 5$.Substituting these values into the formula:$d = \frac{|7 - 5|}{\sqrt{3^2 + (-4)^2}}$$d = \frac{|2|}{\sqrt{9 + 16}}$$d = \frac{2}{\sqrt{25}}$$d = \frac{2}{5}$

Class 11 Mathematics Straight Lines Distance between two lines, Perpendicular distance of the line from a point, Position of point w.r.t. line

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Find the distance between the parallel lines $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$. (in $/5$)

A
$2$
B
$1$
C
$3$
D
$4$

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Distance between two lines, Perpendicular distance of the line from a point, Position of point w.r.t. line

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MediumIIT 1976

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Easy

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EasyMHT CET 2020

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If a line $L$ is perpendicular to the line $5x - y = 1$,and the area of the triangle formed by the line $L$ and the coordinate axes is $5$,then the distance of line $L$ from the line $x + 5y = 0$ is

DifficultJEE MAIN 2014

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Find the distance between the parallel lines $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$. (in $/5$)

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Let the expression $E = 8^a + 8^b - 3 \cdot 2^{a+b}$ take its minimum value $p$ at $a = \alpha$ and $b = \beta$. Then,the perpendicular distance of the point $P(\alpha, \beta)$ from the line $x + y + 2p = 0$ is

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