Hello TomJerry Originally Posted by
TomJerry Find the equation of st. line passing through the point (-2,-3) and inclined at 60 degrees to the line $\displaystyle x+ \sqrt{2} y = 2$
There seems to be a bit of confusion here, so let me see if I can clear it up. Here's a very useful result:If line 1 makes an angle $\displaystyle \theta$ with the positive direction of $\displaystyle Ox$ and line 2 makes an angle $\displaystyle \phi$ with the positive direction of $\displaystyle Ox$, then the angle between line 1 and line 2 is $\displaystyle |\phi-\theta|$.
Suppose that the line we've been given makes an angle $\displaystyle \theta$ with the positive direction of $\displaystyle Ox$, and the line we want makes an angle $\displaystyle \phi$ with $\displaystyle Ox$. Then, since the angle between these two line is $\displaystyle 60^o$:$\displaystyle |\phi-\theta| = 60^o$
In other words:$\displaystyle \phi-\theta = \pm 60^o$
Note that we can choose either the $\displaystyle +$ sign or the $\displaystyle -$ sign. Which one we take will determine whether the $\displaystyle 60^o$ angle is measured clockwise or anticlockwise from the first line. There will therefore be two possible values of $\displaystyle \phi$, and two possible lines.
Next, another fundamental result:If a line makes an angle $\displaystyle \theta$ with the positive direction of $\displaystyle Ox$, its gradient is $\displaystyle \tan \theta$.
The line $\displaystyle x+\sqrt2 y =2$ has a gradient of $\displaystyle -\frac{1}{\sqrt 2}$ (do you agree?), so:$\displaystyle \tan\theta = -\frac{1}{\sqrt 2}$
(By the way, this means that the line makes an angle of about $\displaystyle 35^o$ below the $\displaystyle x$-axis, not $\displaystyle 45^o$.)
So:$\displaystyle \phi-\theta = \pm60^o$
$\displaystyle \Rightarrow \tan(\phi-\theta) = \pm \sqrt 3$
$\displaystyle \Rightarrow \frac{\tan\phi - \tan\theta}{1 + \tan\phi\tan\theta}=\pm\sqrt3$
$\displaystyle \Rightarrow \frac{\tan\phi +\dfrac{1}{\sqrt 2}}{1-\dfrac{1}{\sqrt 2}\tan\phi}=\pm\sqrt3$
$\displaystyle \Rightarrow \frac{\sqrt2\tan\phi+1}{\sqrt2-\tan\phi}=\pm\sqrt3$
$\displaystyle \Rightarrow \sqrt2\tan\phi+1=\pm\sqrt3(\sqrt2-\tan\phi)$
$\displaystyle \Rightarrow \tan\phi(\sqrt2\pm\sqrt3) = \pm\sqrt6-1$
$\displaystyle \Rightarrow \tan\phi = \frac{\sqrt6-1}{\sqrt2+\sqrt3}$ or $\displaystyle \tan\phi = \frac{\sqrt6+1}{\sqrt3-\sqrt2}$
So there are the two gradients that you need. You can now use the equation $\displaystyle y - y_1=m(x-x_1)$ to get the equations of the two possible lines.
(Note: you can check the working here if you like. $\displaystyle \theta = - 35.26^o$. These two values of $\displaystyle \tan\phi$ give $\displaystyle \phi = 24.74^o$ and $\displaystyle 84.74^o$. The first value gives $\displaystyle \phi-\theta = 60^o$; the second gives $\displaystyle \phi-\theta = 120^o$, which is, of course, an angle of $\displaystyle 60^o$ measured in the other direction.)
Grandad