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Is this even possible, given the figure below? I tried to write this line on the form $y=kx+m$, where $k=-\frac{b}{a}.$ Setting $x=0$ I get $y=m$, setting $y=0$ I get $x=\frac{a}{b}m.$ I'm not sure how to proceed here.

The angle $-\theta$ is the angle between the x-axis and the red line, measured clockwise, as shown in the figure.

The answer should be given in terms of $a,b$ and $\theta$.

NOTE: the figure is not really accurate, since I don't know if the red line intercepts the x-axis on the positive side or negative, same for the y-axis. I just drew the problem quickly on geogebra.

enter image description here

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  • $\begingroup$ It can't possibly be an accurate picture since you can't really know where $-b$ is located. You see, depending on what we plug into $-b$, the number might turn out to be positive as well as negative. In other words, $-x$ is a positive number if we substitute it with $-4$ and negative if we substitute it with $4$. The same goes for $a$. $\endgroup$ – Michael Rybkin Nov 6 '17 at 21:28
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Your line passes through the point $(a,-b)$, so it must have the equation $$y = k(x-a) - b$$ where $k$ is the slope of the line. Since the angle of $-\theta$ is marked, you have $$k = \tan(-\theta) = -\tan(\theta),$$ which implies $$ y = -(\tan \theta) (x-a) - b $$

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I don't see why it's not possible. You've got all the necessary information you need to get the line equation for the graph of that line (two points in the plane and the angle which you can use to find the slope of the line). So, the line equation has got to be the following:

$$ y + b = -\tan{\theta}(x - a) $$

The slope here is the tangent of the alternate interior angle at the bottom: $-\theta$. But please don't forget that $\tan{(-\phi)}=-\tan{\phi}$. And, finally, what I used to form this line equation is something called the point-slope form of a line: $y-y_{0}=m(x-x_{0})$.

Now, if you want to know where this line crosses the y- and x-axes, you just need to plug in zero for each of the variables.

y-intercept:

$$ y_{int} + b = -\tan{\theta}(0-a)\implies\\ y_{int} = a\tan{\theta} - b $$

x-intercept: $$ 0 + b = -\tan{\theta}(x_{int}-a)\implies\\ x_{int}=\frac{a\tan{\theta}-b}{\tan{\theta}}\implies\\ x_{int}=a - \frac{b}{\tan{\theta}} $$

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  • $\begingroup$ But can one also express $\tan{\theta}$ in terms of $a$ and $b$? Since $k=-\tan{\theta}=-\frac{b}{a}.$ $\endgroup$ – Parseval Nov 6 '17 at 20:19
  • $\begingroup$ @Parseval no, $\tan \theta \ne b/a$... $\endgroup$ – gt6989b Nov 6 '17 at 20:25
  • $\begingroup$ I don't think so. $\tan{\theta}$ is rise over run. $a$ and $-b$ are not exactly that in your case. $\endgroup$ – Michael Rybkin Nov 6 '17 at 20:26
  • $\begingroup$ In a right triangle, the angle $\tan{\theta}$ is the far side divided by the adjecent side. In my case these sides are -$b$ and $a$. So why doesnt basic trigonometry work here? $\endgroup$ – Parseval Nov 7 '17 at 7:38
  • $\begingroup$ No. $a$ and $-b$ are the distances from the x- and y-axes respectively. They form a point. The slope is rise over run. Rise is the difference in y values ($y_{2}-y_{1}$) and run is the difference in x values ($x_{2}-x_{1}$). You divide them and you get the slope. And that's also the definition of the tangle of an angle. So, to sum up, to find the slope of a straight line, you need two point in the plane or an angle. $\endgroup$ – Michael Rybkin Nov 7 '17 at 9:51

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