The distance between the lines represented by | vedclass

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

Question

(D) The given equation is $x^2+2xy+y^2-8mx-8my-9m^2=0$.This can be rewritten as $(x+y)^2-8m(x+y)-9m^2=0$.Let $X = x+y$. Then the equation becomes $X^2

Vedclass · Accepted Answer

$5\sqrt{2}m$

Vedclass · Answer

(D) The given equation is $x^2+2xy+y^2-8mx-8my-9m^2=0$.This can be rewritten as $(x+y)^2-8m(x+y)-9m^2=0$.Let $X = x+y$. Then the equation becomes $X^2-8mX-9m^2=0$.Factoring the quadratic: $(X-9m)(X+m)=0$.So,$X=9m$ or $X=-m$.This gives two parallel lines: $x+y-9m=0$ and $x+y+m=0$.The distance $d$ between two parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is given by $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.Here,$A=1, B=1, C_1=-9m, C_2=m$.$d = \frac{|-9m-m|}{\sqrt{1^2+1^2}} = \frac{|-10m|}{\sqrt{2}} = \frac{10|m|}{\sqrt{2}} = 5\sqrt{2}|m|$.Assuming $m > 0$,the distance is $5\sqrt{2}m$.

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

The line $x+y=k$ meets the pair of straight lines $x^2+y^2-2x-4y+2=0$ at two points $A$ and $B$. If $O$ is the origin and $\angle AOB=90^{\circ}$,then the value of $k (>1)$ 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 distance between the lines represented by $(x - 2y)^2 + k(x - 2y) = 0$ is $3$ units and $k > 0$, then the equations of the lines are:

If the lines joining the origin to the points of intersection of the curve $2x^2 - 2xy + 3y^2 + 2x - y - 1 = 0$ and the line $x + 2y = k$ are at right angles,then $k^2$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module