0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Solve the two equations and find $x,y$. $\endgroup$ – gambler101 Sep 5 '16 at 12:02
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.