2

A better explanation of my problem

So I'm having a bit of a mathematical dilemma but I'm hoping someone can help me.

I want to know the equation of a line $g(x)$, and it shares an intersection $(9,3)$ with line $y=-0.25x + 5.25$. The intersection makes an angle of $38^\circ$. How might I work out the equation for $g(x)?$ Is there even enough to work with to make this possible?

You can assume that both lines are straight.

Community
  • 1
notHalfBad
  • 121
  • 2
    Given two lines $y=m_1x+n_1$ and $y=m_2x+n_2$, the angle between lines is given by the following formula $$\tan(\alpha)=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$ As you can see there are two options. Find $m_2$, use the given point and find $n_2$. – Galc127 Aug 24 '16 at 05:46
  • so how would I do that in this situation? – notHalfBad Aug 24 '16 at 05:49
  • 1
    You have $\alpha=38^{\circ},m_1=-0.25$. Put the values in formula and get $m_2$. If you need help with solving it, don't hesitate to ask ;) – Galc127 Aug 24 '16 at 06:12
  • Hi @Galc127! I am encountering a similar issue. Can you help me find a point (any) on the line in the direction indicated in the image? – Tanvir Ahmed Feb 01 '23 at 18:32

2 Answers2

1

The easiest way to do this is to use translation-rotation to transform the mobile endpoint first.

See this question (which I answered, so you should know how to prove the result): On the rotation of points issue

That gives you the method of rotating points about the origin. To make that apply here, first translate the endpoints so that the fixed endpoint would "become" the origin. So the mobile endpoint would become $(5-9,4-3) = (-4,1)$

Now apply that counterclockwise rotation transform to this new point:

$(-4,1)$ would become $(-4\cos 38^{\circ} - \sin 38^{\circ}, -4 \sin 38^{\circ} + \cos 38^{\circ})$

And then reverse that linear (translation) transform to get everything back into its rightful place on the plane:

$(-4\cos 38^{\circ} - \sin 38^{\circ}, -4 \sin 38^{\circ} + \cos 38^{\circ})$ would become $(9-4\cos 38^{\circ} - \sin 38^{\circ}, 3-4 \sin 38^{\circ} + \cos 38^{\circ})$

Finally, you have to do is find the equation of the line between that point and $(9,3)$

For that you can simply use $\displaystyle \frac{y-y_1}{x-x_1} = \frac{y- y_2}{x-x_2} $

Community
  • 1
Deepak
  • 26,801
  • Thanks! I'll try that. I'm too young in this site to upvote posts yet but you have a theoretical upvote from me! – notHalfBad Aug 24 '16 at 06:01
  • @notHalfBad I've corrected it. It's a slightly more involved procedure, but it's not that much more difficult. – Deepak Aug 24 '16 at 06:20
0

I haven't the slightest idea.

So I make it up from scratch.

Lines have slopes. Slopes are like angles. How do angles compare to slopes?

A slope is the "rise over the run" i.e. change in distance of the y value over the change in the x value. That has to be proportional to $\sin/\cos$ or $\tan $ of the angle.

So $angle = \arctan slope $.

$Slope = \tan angle $.

That's just common sense. So the angle of $y = -.25x +5.25$ is $\arctan -1/4 = -14$ degrees.

The angle of intersection is $38$ however the question if the intersection is $38$ more than $-14$ or less.

I'll do both. If the line of intersection is $38$ degrees more, The line I want has angle of $38 -14 =24$ so the line has a slope of $\tan 24 =.45$.

The line has point $(9,3) $ and slope $.45$ so the equation of the line is $y -3 = .45 (x -9) $ or $y =.45 x -1.05$

If the angle of intersection is $38$ degrees less, I want an angle of $-14 -38=-52$ or a slope of $\tan _52 = -1.28$

So the equation is $y-3=-1.28 (x-9)$ or $y =-1.28x +14.52$

fleablood
  • 124,253