# Q : 3         Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.                  (iii) $x-y=4$

G Gautam harsolia

Given equation is
$x-y=4$

Coefficient of x is 1 and y is -1
Therefore, $\sqrt{(1)^2+(-1)^2}= \sqrt{1+1}=\sqrt2$
Now, Divide both the sides by $\sqrt2$
we wiil get
$\frac{x}{\sqrt2}-\frac{y}{\sqrt2}= \frac{4}{\sqrt2}$
we can rewrite it as
$x\cos 315\degree + y\sin 315\degree= 2\sqrt2 \ \ \ \ \ \ \ \ \ \ \ -(i)$
Now, we know that normal form of line is
$x\cos \theta + y\sin \theta= p \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Where $\theta$ is the angle between perpendicular and the positive x-axis and p is the perpendicular distance  from the origin
On compairing equation (i) and (ii)
we wiil get
$\theta = 315\degree \ \ and \ \ p = 2\sqrt2$
Therefore,  the angle between perpendicular and the positive x-axis and  perpendicular distance  from the origin is $315\degree \ and \ 2\sqrt2$  respectively

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