There are $5$ to $6$ standard forms of the straight line equation. for example slope intercept form, two intercept form, point slope form and perpendicular form. I have clear visualization of all forms except the perpendicular form. Can any one help me out to give me the concept of perpendicular form. and let me know that how to reduce an equation of straight line for example $x-2y-3=0$ into perpendicular form, i.e. $$x\cos\alpha+y\sin\beta=p$$

  • $\begingroup$ The angles in the cosine and the sine should be the same, so you should replace the $\beta$ with an $\alpha$, as @Narasimham did in his answer. $\endgroup$ Commented Aug 31, 2014 at 22:42

2 Answers 2


Using The Pearson Complete Guide For Aieee 2/e By Khattar as a point of reference for the items below.

You want to convert

$$\tag 1 x-2y-3=0$$

into normal (perpendicular) form.

We have:

$$Ax + By = -C \rightarrow (1)x + (-2)y = -(-3)$$

This means $C \lt 0$, so we divide both sides of $(1)$ by:

$$\sqrt{A^2 + B^2} = \sqrt{(1)^2 + (-2)^2} = \sqrt{5}$$

This yields:

$$\dfrac{1}{\sqrt{5}} x - \dfrac{2}{\sqrt{5}} y = -\dfrac{-3}{\sqrt{5}}$$

In normal form, this is:

$$\cos(\alpha)~x + \sin(\alpha)y = p \rightarrow \dfrac{1}{\sqrt{5}} x - \dfrac{2}{\sqrt{5}} y = -\dfrac{-3}{\sqrt{5}}$$

Note that this has a positive $x$ coordinate, and a negative $y$ coordinate, which puts us in the $4^{th}$ quadrant.

So, we have:

$$\cos(\alpha) = \dfrac{1}{\sqrt{5}}, \sin(-\alpha) = -\dfrac{2}{\sqrt{5}} , p = -\dfrac{-3}{\sqrt{5}} = \dfrac{3}{\sqrt{5}}$$

This gives us:

$$\alpha \approx 1.10714871779409 ~\mbox{radians}~ \approx 63.434948822922 {}^{\circ}$$

Since we are in the $4^{th}$ quadrant, we can write this angle as:

$$\alpha \approx 2 \pi - 1.10714871779409 ~\mbox{radians}~ \approx 5.176036589385497 ~\mbox{radians}~ \approx 296.565 {}^{\circ}$$

As another reference point, see $10.3.4$ (including examples) at Straight Lines.

  • 1
    $\begingroup$ Thank you so much for such a nice description Amzoti.. Now i got it... $\endgroup$
    – zonnie
    Commented Apr 5, 2014 at 9:17
  • $\begingroup$ You should accept an answer that was helpful some time ago. $\endgroup$
    – Narasimham
    Commented Nov 20, 2019 at 11:35

I felt exactly same when first studying the perpendicular form. Seeing components of $p\ \mathrm{as}\ x\cos(\alpha)+y\sin(\alpha)$ line segments by means of a simple construction could burn it into your visual memory: $$p = OL = OQ + QL = OQ + RP =x\cos(\alpha)+y\sin(\alpha)$$

form of straight line


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .