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

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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 $(\omega)$.

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