The joint equation of a pair of straight line | vedclass

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

Question

(A) The slopes of the two lines are $m_1 = 1+\sqrt{2}$ and $m_2 = \frac{1}{1+\sqrt{2}}$.Rationalizing $m_2$: $m_2 = \frac{1}{\sqrt{2}+1} 	imes \frac{

Vedclass · Accepted Answer

$x^2-2 \sqrt{2} x y+y^2=0$

Vedclass · Answer

(A) The slopes of the two lines are $m_1 = 1+\sqrt{2}$ and $m_2 = \frac{1}{1+\sqrt{2}}$.Rationalizing $m_2$: $m_2 = \frac{1}{\sqrt{2}+1} 	imes \frac{\sqrt{2}-1}{\sqrt{2}-1} = \sqrt{2}-1$.The joint equation of lines passing through the origin with slopes $m_1$ and $m_2$ is given by $(y-m_1 x)(y-m_2 x) = 0$,which simplifies to $y^2 - (m_1+m_2)xy + m_1 m_2 x^2 = 0$.Calculate the sum of slopes: $m_1+m_2 = (1+\sqrt{2}) + (\sqrt{2}-1) = 2\sqrt{2}$.Calculate the product of slopes: $m_1 m_2 = (1+\sqrt{2})(\sqrt{2}-1) = (\sqrt{2})^2 - 1^2 = 2-1 = 1$.Substituting these values into the equation: $y^2 - (2\sqrt{2})xy + 1x^2 = 0$,which is $x^2 - 2\sqrt{2}xy + y^2 = 0$.

The joint equation of a pair of straight lines passing through the origin and having slopes $(1+\sqrt{2})$ and $\left(\frac{1}{1+\sqrt{2}}\right)$ is $......$

Similar Questions

The equation $r^2 \cos^2 \left(\theta - \frac{\pi}{3}\right) = 2$ represents

The equation $2x^2 + 4xy - ky^2 + 4x + 2y - 1 = 0$ represents a pair of lines. The value of $k$ is

The joint equation of the lines passing through the origin and trisecting the first quadrant is

The area of the triangle formed by the pair of lines $23x^2 - 48xy + 3y^2 = 0$ with the line $2x + 3y + 5 = 0$ is

If the sum of the slopes of the lines given by $4x^2 + 2\lambda xy - 7y^2 = 0$ is equal to the product of the slopes,then $\lambda$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module