L equiv x cos alpha + y sin alpha - p = 0 rep | vedclass

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

Question

(A) The slope of line $L$ is $-\cot \alpha$. The slope of line $x + y + 1 = 0$ is $-1$. Since the lines are perpendicular,the product of their slopes

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The slope of line $L$ is $-\cot \alpha$. The slope of line $x + y + 1 = 0$ is $-1$. Since the lines are perpendicular,the product of their slopes is $-1$: $(-\cot \alpha)(-1) = -1$ $\Rightarrow \cot \alpha = -1$ $\Rightarrow 	an \alpha = -1$. Since $\alpha$ lies in the fourth quadrant,$\alpha = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$. The perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to $x \cos \alpha + y \sin \alpha - p = 0$ is given by: $\left| \sqrt{2} \cos \alpha + \sqrt{2} \sin \alpha - p ight| = 5$. Substituting $\alpha = \frac{7\pi}{4}$: $\cos \frac{7\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{7\pi}{4} = -\frac{1}{\sqrt{2}}$. $\left| \sqrt{2} \left( \frac{1}{\sqrt{2}} ight) + \sqrt{2} \left( -\frac{1}{\sqrt{2}} ight) - p ight| = 5$. $|1 - 1 - p| = 5 \Rightarrow |-p| = 5$. Since $p$ is positive,$p = 5$.

$L \equiv x \cos \alpha + y \sin \alpha - p = 0$ represents a line perpendicular to the line $x + y + 1 = 0$. If $p$ is positive,$\alpha$ lies in the fourth quadrant,and the perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to the line $L = 0$ is $5$ units,then $p =$

Similar Questions

Find the distance between the parallel lines $15x + 8y - 34 = 0$ and $15x + 8y + 31 = 0$.

The line $L$ is given by $\frac{x}{5} + \frac{y}{b} = 1$ and passes through the point $(13, 32)$. The line $K$ is parallel to $L$ and has the equation $\frac{x}{c} + \frac{y}{3} = 1$. Then the distance between $L$ and $K$ is

If the perpendicular distances from the points $(2, 3)$,$(4, a)$ and $(\alpha, \beta)$ to the line $3x + 4y - 3 = 0$ are equal and $4\alpha - 3\beta + 1 = 0$,then the sum of all possible values of $a$,$\alpha$,and $\beta$ is:

The number of points on the line $x+y=4$ which are at a unit distance from the line $2x+2y=5$ is

The length of the perpendicular from the point $(3, 1)$ to the line $4x + 3y + 20 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module