A straight line passes through the point $P(2,√3)$ and makes an angle of $60°$ with X- axis. If this line intersects another line having equation $x+y√3 =12$ at $Q$, find the length of $PQ$.

My Attempt:

Here, the slope of the line passing through $P$ and making an angle of $60°$ with X- axis is: $$m=Tan60°$$ $$m=√3$$ Then,

Its equation is $$y-y_1=m(x-x_1)$$ $$y-√3 = √3 (x-2)$$ $$y-√3 = √3 x-2√3$$ $$√3 x - y - √3 =0$$

I could not continue from here. Please help to complete it.

  • $\begingroup$ Solve the two equations and find $x,y$. $\endgroup$ – gambler101 Sep 5 '16 at 12:02

You're doing well so far.

You have two lines now:

$\sqrt3 x - y - \sqrt3 =0$ and $x+y\sqrt3 =12$

You need to solve this pair of simultaneous equations to find $Q$

I would probably rearrange the first to make it $y=\sqrt3x-\sqrt3$

Then substitute that into the second:

$x+y\sqrt3 =12 \Rightarrow x+(\sqrt3x-\sqrt3)\sqrt3 =12$

$x+3x-3 =12$

and continue to find $x$ and $y$ etc.

  • $\begingroup$ I got as $$x=\frac {15}{4}$$ and $$y=\frac {11√3}{4}$$. Is this correct? $\endgroup$ – pi-π Sep 5 '16 at 12:07
  • $\begingroup$ Looks good to me. You can check by substituting your values into the equations. For example, $x+y\sqrt3=\frac{15}4 + \frac{11\sqrt3 \sqrt3}4=\frac{15+33}4=\frac{48}4=12$ as required. $\endgroup$ – tomi Sep 6 '16 at 22:42
  • $\begingroup$ Now you need to find the distance between $P(2,\sqrt3)$ and $Q(\frac{15}4,\frac{11\sqrt3}4)$ $\endgroup$ – tomi Sep 6 '16 at 22:44

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