(A) The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.Giv

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.Given lines are $3x - 2y + 5 = 0$ and $3x - 2y + C = 0$.Here,$a = 3$,$b = -2$,$c_1 = 5$,$c_2 = C$,and $d = 5$.Substituting these values into the formula:$5 = \frac{|C - 5|}{\sqrt{3^2 + (-2)^2}}$$5 = \frac{|C - 5|}{\sqrt{9 + 4}}$$5 = \frac{|C - 5|}{\sqrt{13}}$$|C - 5| = 5\sqrt{13}$$C - 5 = \pm 5\sqrt{13}$$C = 5 \pm 5\sqrt{13} = 5(1 \pm \sqrt{13})$.

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

Easy

If the equation of a line parallel to $3x - 2y + 5 = 0$ and at a distance of $5$ units from it is $3x - 2y + C = 0$,then $C$ is equal to

TS EAMCET 2021 All TS EAMCET

A
$5(1 \pm \sqrt{13})$
B
$5(\pm \sqrt{13} + 1)$
C
$5(\sqrt{13} \pm 1)$
D
$5(\frac{-1 \pm \sqrt{13}}{\sqrt{13}})$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

Straight Line

Subtopic

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

Previous Year

TS EAMCET 2021 Paper All TS EAMCET years →

Find the two points on the line $x + y = 4$ which are at a unit distance from the line $4x + 3y = 10$.

Difficult

View Solution

If $p$ and $q$ are lengths of the perpendiculars from the origin to the lines $x \sec \theta + y \operatorname{cosec} \theta = k$ and $x \cos \theta - y \sin \theta = k \cos 2\theta$ respectively,then

MediumAP EAMCET 2020

View Solution

If the point $(a, 2a)$ lies between the lines $|x + y + 1| = 4$,then the complete set of values of $'a'$ is:

View Solution

The length of the perpendicular from the point $(a \cos \alpha, a \sin \alpha)$ upon the straight line $y = x \tan \alpha + c$,where $c > 0$,is:

Easy

View Solution

The points on the $x$-axis whose perpendicular distance from the line $\frac{x}{a} + \frac{y}{b} = 1$ is $a$ are

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

If the equation of a line parallel to $3x - 2y + 5 = 0$ and at a distance of $5$ units from it is $3x - 2y + C = 0$,then $C$ is equal to

Similar Questions

Find the two points on the line $x + y = 4$ which are at a unit distance from the line $4x + 3y = 10$.

If $p$ and $q$ are lengths of the perpendiculars from the origin to the lines $x \sec \theta + y \operatorname{cosec} \theta = k$ and $x \cos \theta - y \sin \theta = k \cos 2\theta$ respectively,then

If the point $(a, 2a)$ lies between the lines $|x + y + 1| = 4$,then the complete set of values of $'a'$ is:

The length of the perpendicular from the point $(a \cos \alpha, a \sin \alpha)$ upon the straight line $y = x \tan \alpha + c$,where $c > 0$,is:

The points on the $x$-axis whose perpendicular distance from the line $\frac{x}{a} + \frac{y}{b} = 1$ is $a$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module