The slope of a line L passing through the poi | vedclass

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

Question

(A) The line $L$ passes through $(-2, -3)$ and has an undefined slope,so its equation is $x = -2$ (a vertical line).The angle between $x = -2$ and the

Vedclass · Accepted Answer

$\pi - 	an^{-1}\left(\frac{1}{2}ight)$

Vedclass · Answer

(A) The line $L$ passes through $(-2, -3)$ and has an undefined slope,so its equation is $x = -2$ (a vertical line).The angle between $x = -2$ and the line $ax - 2y + 3 = 0$ is $45^{\circ}$.The slope of the line $ax - 2y + 3 = 0$ is $m_1 = \frac{a}{2}$.The angle $	heta$ between a vertical line and a line with slope $m_1$ is given by $|	an(90^{\circ} - 	heta)| = |\frac{1}{m_1}|$.Since the angle is $45^{\circ}$,we have $|\frac{1}{a/2}| = 	an(45^{\circ}) = 1$.Thus,$\frac{2}{a} = 1$,which gives $a = 2$.Now,substitute $a = 2$ into the equation $x + ay - 4 = 0$ to get $x + 2y - 4 = 0$.The slope of this line is $m = -\frac{1}{2}$.The angle $	heta$ made with the positive $X$-axis is given by $	an 	heta = -\frac{1}{2}$.Since the slope is negative,the angle is in the second quadrant,so $	heta = \pi - 	an^{-1}\left(\frac{1}{2}ight)$.

The slope of a line $L$ passing through the point $(-2, -3)$ is not defined. If the angle between the lines $L$ and $ax - 2y + 3 = 0$ $(a > 0)$ is $45^{\circ}$,then the angle made by the line $x + ay - 4 = 0$ with the positive $X$-axis in the anticlockwise direction is

Similar Questions

The lines $p(p^2 + 1)x - y + q = 0$ and $(p^2 + 1)^2x + (p^2 + 1)y + 2q = 0$ are perpendicular to a common line for :

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

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

The equation of one of the straight lines which passes through the point $(1, 3)$ and makes an angle of $\tan^{-1}(\sqrt{2})$ with the straight line $y + 1 = 3\sqrt{2}x$ is:

If the equation of a line having a slope $m$ $(m \in \mathbb{Z})$,passing through $(1, 1)$ and making an angle of $\tan^{-1}\left(\frac{5}{7}\right)$ with the line $x + y - 3 = 0$ is $ax + y + c = 0$,then $ac =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module