# Determining the equation of line using the Normal form.

A straight line makes an intercept on the y axis twice as long as that on the x axis and is at a unit distance from the origin. Determine it's equation.

Consider two points on the line $(0,2a)$ and $(a,0)$

$\tan\omega = (0-2a)/(a-o)= -2$ // where $\omega$ is the angle the line makes with the positive direction of X axis

Therefore, $\sin \omega = \dfrac{-2}{\sqrt5}$ and $\cos\omega= \dfrac{1}{\sqrt5}$

Now,

Normal of straight line equation is:

$x\cos\omega+y\sin\omega = |p|$ where |p| is the distance from the origin.

Thus, I obtained two equations:

$x-2y = \pm\sqrt5$

However, the answer given in the key is $2x+y= \pm\sqrt5$

I can't find my mistake. Where have I gone wrong?

• What do you denote ω? The polar angle of the straight line? – Bernard Aug 7 '17 at 13:00
• @Bernard $\omega$ is the angle the line makes with the positive direction of X axis – user342531 Aug 7 '17 at 13:02

Look at the diagram below:

Note that the area of the right triangle $ABO$ is: $$S_{\Delta ABO}=\frac{1}{2}AO\cdot BO=\frac{1}{2}\cdot 2a\cdot a=a^2$$ $$S_{\Delta ABO}=\frac{1}{2}AB\cdot OC=\frac{1}{2}\cdot a\sqrt{5}\cdot 1=\frac{a\sqrt{5}}{2}.$$ Equate them: $$a=\frac{\sqrt{5}}{2}.$$

The equation of the line: $$\frac{y-2a}{0-2a}=\frac{x-0}{a-0} \Rightarrow y=-2x+2a.$$

Note that there are four symmetrical cases of the diagram above. Can you figure out other cases?

HINT: the equation of the line is $$y=-2x+2a$$ converting this in the Hessian Normalform you can compute $a$ set $$y=mx+n$$ since $$P_1(0;2a)$$ and $$P_2(a;0)$$ is given we can compute $$m=\frac{2a-0}{-a}=-2$$ and $$n=2a$$ the Hessian Normalform is given by $$\frac{2x+y-2a}{\pm\sqrt{5}}=0$$ if you plug in $$x=0,y=0$$ then you can derive $a$

• Please explain how you arrived at that equation and please point out my mistake. – user342531 Aug 7 '17 at 13:12
• what is this here? $-1$ Why? – Dr. Sonnhard Graubner Aug 7 '17 at 13:12

The error comes from the value of $\sin \omega$: you forgot the polar angle of a straight line, as opposed to the polar angle of a vector, belongs to $\mathbf R/\pi\mathbf Z$ – in other words, it is represented by an element in $[0,\pi)$, and consequently $\sin\omega\ge 0$.

That said, you didn't use the normal form of the equation: if the $x$-intercept of the straight line is $a$ and its $y$-intercept is $b$ (both non-zero), the equation of the line is $$\frac x a+\frac yb =1.$$

Let's plug in the actual values: the equation is $$\frac xa+\frac y{2a}=1,$$ and a parametric equation of the normal from the origin to the line is $$x=\frac ta,\quad y=\frac t{2a}\qquad(t\in\mathbf R).$$ The intersection point $H$ of this normal to the straight line corresponds to the value of $t$ such that the equation of the line is satisfied: $$\frac t{a^2}+\frac t{4a^2}=1,\enspace\text{whence}\quad t=\frac{4a^2}5,\;x_H=\frac{4a}5,\; y_H=\frac{2a}5$$ and the distance from the origin to the straight line is $$OH=\sqrt{x_H^2+y_H^2}=\frac{2\lvert a\rvert}{\sqrt 5}=1\iff \lvert a\rvert=\frac{\sqrt 5}2,$$ whence the equation in normal form: $$\frac{2\sqrt 5}5x+\frac{\sqrt 5}5y=\pm 1.$$