(C) The given equation is $x^2+2 \sqrt{2} xy+2y^2+4x+4 \sqrt{2}y+1=0$. Comparing this with the general equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a

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

(C) The given equation is $x^2+2 \sqrt{2} xy+2y^2+4x+4 \sqrt{2}y+1=0$. Comparing this with the general equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1$,$h=\sqrt{2}$,$b=2$,$g=2$,$f=2 \sqrt{2}$,and $c=1$. For parallel lines,the distance $d$ is given by the formula $d=2 \sqrt{\frac{g^2-ac}{a(a+b)}}$. Substituting the values: $d=2 \sqrt{\frac{2^2-(1)(1)}{1(1+2)}}$. $d=2 \sqrt{\frac{4-1}{3}} = 2 \sqrt{\frac{3}{3}} = 2 \sqrt{1} = 2$.

Class 11 Mathematics Pair of straight lines Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

Easy

The distance between the pair of lines $x^2+2 \sqrt{2} xy+2y^2+4x+4 \sqrt{2}y+1=0$ is

AP EAMCET 2019 All AP EAMCET

A
$4 \sqrt{2}$
B
$2 \sqrt{2}$
C
$2$
D
$6 \sqrt{2}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

Pair of straight lines

Subtopic

Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

Previous Year

AP EAMCET 2019 Paper All AP EAMCET years →

If the distance between the parallel lines given by the equation $x^2+4xy+4y^2+3x+6y-4=0$ is $\lambda$,then $\lambda^2=$

MediumMHT CET 2023

View Solution

If the lines joining the origin to the points of intersection of $2 x+3 y=k$ and $3 x^2-x y+3 y^2+2 x-3 y-4=0$ are at right angles,then

MediumAP EAMCET 2023

View Solution

The distance between the parallel lines represented by the equation $x^2+2xy+y^2-8mx-8my-9m^2=0$ is

DifficultAP EAMCET 2017

View Solution

If the equation $x^2+2 \sqrt{2} xy + 2y^2 + 4x + 4 \sqrt{2}y + 1 = 0$ represents a pair of parallel straight lines,find the distance between them.

MediumAP EAMCET 2020

View Solution

If the combined equation of the lines joining the origin to the points of intersection of the curve $x^2+y^2-2x-4y+2=0$ and the line $x+y-2=0$ is $(l_1x+m_1y)(l_2x+m_2y)=0$,then $l_1+l_2+m_1+m_2=$

MediumAP EAMCET 2025

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

The distance between the pair of lines $x^2+2 \sqrt{2} xy+2y^2+4x+4 \sqrt{2}y+1=0$ is

Similar Questions

If the distance between the parallel lines given by the equation $x^2+4xy+4y^2+3x+6y-4=0$ is $\lambda$,then $\lambda^2=$

If the lines joining the origin to the points of intersection of $2 x+3 y=k$ and $3 x^2-x y+3 y^2+2 x-3 y-4=0$ are at right angles,then

The distance between the parallel lines represented by the equation $x^2+2xy+y^2-8mx-8my-9m^2=0$ is

If the equation $x^2+2 \sqrt{2} xy + 2y^2 + 4x + 4 \sqrt{2}y + 1 = 0$ represents a pair of parallel straight lines,find the distance between them.

If the combined equation of the lines joining the origin to the points of intersection of the curve $x^2+y^2-2x-4y+2=0$ and the line $x+y-2=0$ is $(l_1x+m_1y)(l_2x+m_2y)=0$,then $l_1+l_2+m_1+m_2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module