A line passes through the origin and makes eq | vedclass

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

Question

(D) The line passing through the origin and making equal angles with the positive coordinate axes is $y = x$.$L_1: 2x + y + 6 = 0$ and $L_2: 4x + 2y -

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) The line passing through the origin and making equal angles with the positive coordinate axes is $y = x$.$L_1: 2x + y + 6 = 0$ and $L_2: 4x + 2y - p = 0$ (or $2x + y - \frac{p}{2} = 0$).Since $L_1$ and $L_2$ have the same slope $m = -2$,they are parallel lines.The line $y = x$ intersects $L_1$ at $A$ and $L_2$ at $B$.In the right-angled triangle $	riangle AMB$,the angle $	heta$ between the line $y = x$ (slope $m_1 = 1$) and the line $L_2$ (slope $m_2 = -2$) is given by:$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \left| \frac{1 - (-2)}{1 + (1)(-2)} ight| = \left| \frac{3}{-1} ight| = 3$.In $	riangle AMB$,$	an 	heta = \frac{AM}{BM}$.Therefore,$\frac{AM}{BM} = 3$.

Similar Questions

Show that the equation of the line passing through the origin and making an angle $\theta$ with the line $y=mx+c$ is $\frac{y}{x}=\frac{m \pm \tan \theta}{1 \mp m \tan \theta}$.

The equations of the lines passing through the point $(3,2)$ which make an angle of $45^{\circ}$ with the line $x-2y=3$ are

What is the angle in degrees between the lines $y = x + 5$ and $y = \sqrt{3}x - 4$ (in $^\circ$)?

The angle between the lines $2x - y - 15 = 0$ and $3x + y + 4 = 0$ is .....$^o$.

$A$ straight line $L$ through the point $(3,-2)$ is inclined at an angle of $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If $L$ also intersects the $X$-axis,then the equation of $L$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module