How do I determine k so that the line of the beam is parallel to a $60^\circ$ angle?

Question

I have the equation of a beam that looks like this: $$(x + y - 5) + k(2x - 3y) = 0$$

I know that the angular coefficient of a $60^\circ$ angle is equivalent to the root of 3.

$$m = \sqrt3$$

Though, how do I determine k so that the line of the beam is parallel a 60 degrees angle with the x-axis?

I don't remember, how do grads or grades compare with degrees? — Will Jagy, Mar 04 '15 at 20:45
Also, $\sqrt 3$ is associated with $60^\circ,$ so i think there is an error here. — Will Jagy, Mar 04 '15 at 20:46
I just edited my question. I'm very sorry. I had to translate my problem into English and I missed some bits. — Cesare, Mar 04 '15 at 20:47
I double checked, the angular coefficient, $m$, of $60^\circ$ seems to be equivalent to $\sqrt 3$. I appreciate your help. — Cesare, Mar 04 '15 at 20:52

score 1 · Accepted Answer · answered Mar 04 '15 at 20:57

1

The given equation is $(2k+1)x+(1-3k)y-5=0$. The slope of the beam is ${2k+1\over 3k-1}$ which must be equal to $\sqrt 3$. That is $${2k+1\over 3k-1}=\sqrt 3$$ Can you continue and find $k$?

answered Mar 04 '15 at 20:57

Fermat

5,230

Thank you very much for answering my question! All of this sounds great. Though, how did you find the slope of the beam? Isn't it $2k+1\over 1-3k$? Thanks again. – Cesare Mar 04 '15 at 21:05
Did you basically find the $m$ of the beam using this formula: -$a\over b$ of which $a$ and $b$ are the coefficient of the beam $ax + by + c$? – Cesare Mar 04 '15 at 21:11
Yes. The slope of the line $ax+by+c=0$ is $-{a\over b}$. @CeceXX – Fermat Mar 04 '15 at 21:22

How do I determine k so that the line of the beam is parallel to a $60^\circ$ angle?

1 Answers1