What is the distance between the lines repres | vedclass

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

Question

(C) The given equation can be rewritten as $x^2 + 2(\sqrt{2}y + 2)x + (2y^2 + 4\sqrt{2}y + 1) = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The given equation can be rewritten as $x^2 + 2(\sqrt{2}y + 2)x + (2y^2 + 4\sqrt{2}y + 1) = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:$x = \frac{-2(\sqrt{2}y + 2) \pm \sqrt{4(\sqrt{2}y + 2)^2 - 4(2y^2 + 4\sqrt{2}y + 1)}}{2}$$x = -(\sqrt{2}y + 2) \pm \sqrt{(\sqrt{2}y + 2)^2 - (2y^2 + 4\sqrt{2}y + 1)}$$x = -\sqrt{2}y - 2 \pm \sqrt{2y^2 + 4\sqrt{2}y + 4 - 2y^2 - 4\sqrt{2}y - 1}$$x = -\sqrt{2}y - 2 \pm \sqrt{3}$.Thus,the lines are $x + \sqrt{2}y + 2 - \sqrt{3} = 0$ and $x + \sqrt{2}y + 2 + \sqrt{3} = 0$.These are parallel lines of the form $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$.The distance between them is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.$d = \frac{|(2 + \sqrt{3}) - (2 - \sqrt{3})|}{\sqrt{1^2 + (\sqrt{2})^2}} = \frac{|2\sqrt{3}|}{\sqrt{3}} = 2$.

What is the distance between the lines represented by the equation $x^2 + 2\sqrt{2}xy + 2y^2 + 4x + 4\sqrt{2}y + 1 = 0$?

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

If the pair of lines joining the origin to the points of intersection of the line $x+y=1$ with the curve $x^2+y^2+2hxy+gx+fy+1=0$ are at right angles,then the point $(g, f)$ lies on the line

$A$ pair of lines drawn through the origin forms a right-angled isosceles triangle with the line $2x + 3y = 6$,having the right angle at the origin. The area (in sq. units) of the triangle thus formed is

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 $l_1+l_2+m_1+m_2=$

Find the equation of the line passing through the points $(-1, 1)$ and $(2, -4)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module