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Q3: 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

Answers (1)

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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 sides by \sqrt2
we will 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 the perpendicular and the positive x-axis, and p is the perpendicular distance from the origin?
On comparing equation (i) and (ii)
we will get
\theta = 315\degree \ \ and \ \ p = 2\sqrt2
Therefore,  the angle between perpendicular and the positive x-axis and  perpendicular distance from the origin are 315\degree \ and \ 2\sqrt2  respectively

Posted by

Gautam harsolia

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