The joint equation of the pair of lines passi | vedclass

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

Question

(C) 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{\sqrt{2}-1}{(\sqrt{2}+1)(\sq

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The slopes of the two lines are $m_1 = 1+\sqrt{2}$ and $m_2 = \frac{1}{1+\sqrt{2}}$.Rationalizing $m_2$: $m_2 = \frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1$.The equations of the lines passing through the origin are $y = m_1x$ and $y = m_2x$,which are $y = (1+\sqrt{2})x$ and $y = (\sqrt{2}-1)x$.Rearranging these gives $(1+\sqrt{2})x - y = 0$ and $(\sqrt{2}-1)x - y = 0$.The joint equation is given by the product: $[(1+\sqrt{2})x - y][(\sqrt{2}-1)x - y] = 0$.Expanding this: $(1+\sqrt{2})(\sqrt{2}-1)x^2 - (1+\sqrt{2})xy - (\sqrt{2}-1)xy + y^2 = 0$.$(2-1)x^2 - (1+\sqrt{2}+\sqrt{2}-1)xy + y^2 = 0$.$x^2 - 2\sqrt{2}xy + y^2 = 0$.

The joint equation of the pair of lines passing through the origin with slopes $(1+\sqrt{2})$ and $\frac{1}{(1+\sqrt{2})}$ is

Similar Questions

If the equation $\lambda x^{2} + 2y^{2} - 5xy + 5x - 7y + 3 = 0$ represents a pair of straight lines,then $\lambda = \dots$

The value of $\lambda$, for which the equation $x^2 - y^2 - x - \lambda y - 2 = 0$ represents a pair of straight lines, is

The slopes of the lines represented by $6x^2 + 2hxy + y^2 = 0$ are in the ratio $2:3$,then $h =$

If the sum and product of the slopes of the lines represented by $4x^2 + 2hxy - 7y^2 = 0$ are equal,then $h = \dots$

The combined equation of the straight lines of the form $y=kx+1$ (where $k$ is an integer) such that the point of intersection of each with the line $3x+4y=9$ has an integer as its $x$-coordinate is

Vedclass Test Series

Exam Paper Generator

Online Exam Module