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

The lines joining the points of intersection of the curve $(x - h)^2 + (y - k)^2 - c^2 = 0$ and the line $kx + hy = 2hk$ to the origin are perpendicular,then

$A$ pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve $x^2+y^2=4$ with $x+y=a$. The set containing the value of $a$ is

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

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.

The area (in sq. units) of the triangle formed by the lines $x^2-3xy+y^2=0$ and $x+y+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module