# Math Help - Find the eq of st line

1. ## Find the eq of st line

Find the equation of st. line passing through the point (-2,-3) and inclined at 60 degrees to the line $x+ \sqrt{2} y = 2$

2. Dear TomJerry,

1)Draw the given line and the point as a rough sketch.

2)As the given line makes an ange of 45 degrees with the x axis I think by drawing the line and the point you will be able to think of the tangent of the line that you have to find.

But if you need futher assistance please don't hesitate to reply back.

Hope this helps.

3. I have already done plotting the graph and only after that when i wasnt able to device the eq had posted this question on the forum to u guys......

My work ..............what i have done till now

Eq of a line $(y-y_0) = m (x-x_0)$

m is the slope

$m = tan^{-1} \left(\frac{\pi}{3}\right)$

$m = \sqrt{3}$

Eq of line is $x+\sqrt{3} = 2 \Rightarrow y = \frac{1}{\sqrt{3}}(2-x)$ here $m = \frac{1}{\sqrt{3}}$

$tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

$\sqrt{3} = \left| \frac{m_1 - \frac{1}{\sqrt{3}}}{1 + m_1 \frac{1}{\sqrt{3}}} \right|$

squaring both the terms
$m^2 = \sqrt{\frac{5}{3}}$

equation of line is
$y+3 = \sqrt{\frac{5}{3}} (x+2)$ ......................Is this correct ???????????

4. Dear TomJerry,

Please reconsider the value you have taken to the slope (m) of the line. According to the question relative to what the slope is given?
Yes you are correct that the question asks only to find the equation not to draw the line. But on the contart since straight lines could be easily drawn and since the questionis about straight lines if you know how to sketch a straight line it would be helpful.

5. Originally Posted by Sudharaka
Dear TomJerry,

Please reconsider the value you have taken to the slope (m) of the line. According to the question relative to what the slope is given?
Yes you are correct that the question asks only to find the equation not to draw the line. But on the contart since straight lines could be easily drawn and since the questionis about straight lines if you know how to sketch a straight line it would be helpful.
thanks but could u please tell me what is wrng with my result .so that i can rectify it ......

6. ## Angle between two lines in coordinate geometry

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 $x+ \sqrt{2} y = 2$
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 $\theta$ with the positive direction of $Ox$ and line 2 makes an angle $\phi$ with the positive direction of $Ox$, then the angle between line 1 and line 2 is $|\phi-\theta|$.
Suppose that the line we've been given makes an angle $\theta$ with the positive direction of $Ox$, and the line we want makes an angle $\phi$ with $Ox$. Then, since the angle between these two line is $60^o$:
$|\phi-\theta| = 60^o$
In other words:
$\phi-\theta = \pm 60^o$
Note that we can choose either the $+$ sign or the $-$ sign. Which one we take will determine whether the $60^o$ angle is measured clockwise or anticlockwise from the first line. There will therefore be two possible values of $\phi$, and two possible lines.

Next, another fundamental result:
If a line makes an angle $\theta$ with the positive direction of $Ox$, its gradient is $\tan \theta$.
The line $x+\sqrt2 y =2$ has a gradient of $-\frac{1}{\sqrt 2}$ (do you agree?), so:
$\tan\theta = -\frac{1}{\sqrt 2}$
(By the way, this means that the line makes an angle of about $35^o$ below the $x$-axis, not $45^o$.)

So:
$\phi-\theta = \pm60^o$

$\Rightarrow \tan(\phi-\theta) = \pm \sqrt 3$

$\Rightarrow \frac{\tan\phi - \tan\theta}{1 + \tan\phi\tan\theta}=\pm\sqrt3$

$\Rightarrow \frac{\tan\phi +\dfrac{1}{\sqrt 2}}{1-\dfrac{1}{\sqrt 2}\tan\phi}=\pm\sqrt3$

$\Rightarrow \frac{\sqrt2\tan\phi+1}{\sqrt2-\tan\phi}=\pm\sqrt3$

$\Rightarrow \sqrt2\tan\phi+1=\pm\sqrt3(\sqrt2-\tan\phi)$

$\Rightarrow \tan\phi(\sqrt2\pm\sqrt3) = \pm\sqrt6-1$

$\Rightarrow \tan\phi = \frac{\sqrt6-1}{\sqrt2+\sqrt3}$ or $\tan\phi = \frac{\sqrt6+1}{\sqrt3-\sqrt2}$
So there are the two gradients that you need. You can now use the equation $y - y_1=m(x-x_1)$ to get the equations of the two possible lines.

(Note: you can check the working here if you like. $\theta = - 35.26^o$. These two values of $\tan\phi$ give $\phi = 24.74^o$ and $84.74^o$. The first value gives $\phi-\theta = 60^o$; the second gives $\phi-\theta = 120^o$, which is, of course, an angle of $60^o$ measured in the other direction.)