If the normal drawn from the origin to the st | vedclass

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

Question

(C) The equation of the line is $2x + 7y + 6 = 0$.Slope of the line $m = -\frac{2}{7}$.The normal drawn from the origin to the line is perpendicular t

Vedclass · Accepted Answer

$\pi + 	an^{-1} \frac{7}{2}$

Vedclass · Answer

(C) The equation of the line is $2x + 7y + 6 = 0$.Slope of the line $m = -\frac{2}{7}$.The normal drawn from the origin to the line is perpendicular to the line.Let the slope of the normal be $m'$. Since the normal is perpendicular to the line,$m 	imes m' = -1$.$(-\frac{2}{7}) 	imes m' = -1 \Rightarrow m' = \frac{7}{2}$.Let $\alpha$ be the angle the normal makes with the positive $X$-axis. Then $	an \alpha = m' = \frac{7}{2}$,so $\alpha = 	an^{-1} \frac{7}{2}$.From the figure,the normal lies in the third quadrant,so the angle $	heta$ with the positive $X$-axis is $\pi + \alpha$.Therefore,$	heta = \pi + 	an^{-1} \frac{7}{2}$.

If the normal drawn from the origin to the straight line $2x + 7y + 6 = 0$ makes an angle $\theta$ with the positive $X$-axis,then $\theta =$

Similar Questions

If $(h, k)$ is the image of the point $(2, -3)$ with respect to the line $5x - 3y = 2$,then $h + k =$

If $2x + 3y = 5$ is the perpendicular bisector of the line segment joining the points $A\left(1, \frac{1}{3}\right)$ and $B$,then $B$ is equal to

The coordinates of the foot of the perpendicular drawn from $(0,0)$ to the line joining $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$ are

Let $M$ be the foot of the perpendicular drawn from the point $(5, -7)$ to the line $3x - 5y + 1 = 0$. Then the perpendicular distance from $M$ to the line $2x + 5y - 3 = 0$ is

If $(\alpha, \beta)$ is the image of the point $(3, -4)$ with respect to the line $4x - y - 1 = 0$,then the value of $\beta - \alpha$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module