Reduce the equation x-y=4 into the normal for | vedclass

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

Question

(A) The given equation is $x - y = 4$.To reduce it to the normal form $x \cos \omega + y \sin \omega = p$,we divide the equation by $\sqrt{A^2 + B^2}$

Vedclass · Accepted Answer

$p = 2\sqrt{2}, \omega = 315^{\circ}$

Vedclass · Answer

(A) The given equation is $x - y = 4$.To reduce it to the normal form $x \cos \omega + y \sin \omega = p$,we divide the equation by $\sqrt{A^2 + B^2}$,where $A = 1$ and $B = -1$.$\sqrt{1^2 + (-1)^2} = \sqrt{2}$.Dividing by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y = \frac{4}{\sqrt{2}}$.This simplifies to $\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y = 2\sqrt{2}$.Comparing this with $x \cos \omega + y \sin \omega = p$,we have $\cos \omega = \frac{1}{\sqrt{2}}$ and $\sin \omega = -\frac{1}{\sqrt{2}}$.Since $\cos \omega > 0$ and $\sin \omega $\omega = 360^{\circ} - 45^{\circ} = 315^{\circ}$.The perpendicular distance $p = 2\sqrt{2}$ and the angle $\omega = 315^{\circ}$.

Reduce the equation $x-y=4$ into the normal form $x \cos \omega + y \sin \omega = p$. Find the perpendicular distance from the origin $(p)$ and the angle between the perpendicular and the positive $x$-axis $(\omega)$.

Similar Questions

The distance between the two parallel lines $y = 2x + 7$ and $y = 2x + 5$ is:

The line $L_1$ given by $\frac{x}{p} + \frac{y}{2} = 1$ passes through the point $(5, 0)$. The line $L_2$ given by $\frac{x}{10} + \frac{y}{q} = 1$ is parallel to $L_1$. Find the distance between the lines $L_1$ and $L_2$.

$A$ line has slope $m$ and $y$-intercept $4$. The distance between the origin and the line is equal to

$A$ straight line makes equal negative intercepts on the coordinate axes. If the perpendicular distance from the origin to the line is $4 \text{ units}$, then the equation of the line is:

$P$ is a point on $x+y+5=0$,whose perpendicular distance from $2x+3y+3=0$ is $\sqrt{13}$. Then the coordinates of $P$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module