Concurrency of lines

Question

If the three lines: $$x\sin^2 \theta + y \cos^2 \theta = 1$$ $$x \cos^2 \theta + y \sin^2 \theta = 1$$ $$lx + my + n = 0$$ are concurrent then which of the following is true?

a) $l+m=n$

b) $l^3+m^3+n^3=3lmn$

c) $l=m=n$

d) None of these

Well obviously if the lines are concurrent, then all three terms have to be in a common ratio. However don't know where to proceed from there. Please help!

Note that we have $l^3+m^3+n^3=3lmn+(l+m+n)(l^2+m^2+n^2-lm-mn-nl)$ which is a cubic identity worth knowing. — Mark Bennet, Aug 30 '14 at 16:47
@MarkBennet Ummmm.... lost you there. What exactly do you mean? — Gummy bears, Aug 30 '14 at 16:47
Yeah lol I realized that. Thanks though! I can definitely pester my teacher about this one :D — Gummy bears, Aug 30 '14 at 17:12

score 1 · Answer 1 · answered Aug 30 '14 at 16:32

1

Solving for $x,y$ from the first two equation, we find $x=y=1$

So for concurrency, $lx+my+n=0$ must pass through $(1,1)$

answered Aug 30 '14 at 16:32

lab bhattacharjee

274,582

Hmmm... How exactly did you solve for $x,y$? – Gummy bears Aug 30 '14 at 16:33
@Gummybears, Do you know how to solve for two unknowns from two linear simultaneous equation – lab bhattacharjee Aug 30 '14 at 16:36
Yes of course. But in this it'll be kinda hard o.O – Gummy bears Aug 30 '14 at 16:36
Note: Except for the case where $\sin \theta = \pm \cos \theta$ when $x+y=2$ and the two first lines are identical – Mark Bennet Aug 30 '14 at 16:41
@MarkBennet, I was going to add. I think the question should have included that implicit condition – lab bhattacharjee Aug 30 '14 at 16:42
I agree it is an odd question otherwise. – Mark Bennet Aug 30 '14 at 16:45
Well it has the option of none of these :P – Gummy bears Aug 30 '14 at 16:46
@Gummybears, Then we should have to write Option $$(b)$$ if $\cos2\theta\ne0$ $$(d)$$ otherwise – lab bhattacharjee Aug 30 '14 at 16:48
Oh? Ummm... Wiat I think I lost you there. How exactly is (b) true? – Gummy bears Aug 30 '14 at 16:49
@Gummybears, http://math.stackexchange.com/questions/543991/factorize-the-polynomial-x3y3z3-3xyz?lq=1 – lab bhattacharjee Aug 30 '14 at 16:56

score 0 · Answer 2 · answered Aug 30 '14 at 16:35

0

The three lines are concurrent iff the determinant of $$ \left[ \begin{array}{ccc} s^2& c^2& -1\\ c^2& s^2& -1\\ l& m& n \end{array} \right] $$ is equal to zero. You shall have $l+m+n=0$ or $s^2-c^2=0$.

answered Aug 30 '14 at 16:35

Troy Woo

3,579

Ohhh..... Well that's a good thing to know. – Gummy bears Aug 30 '14 at 16:37
1

@Gummybears, http://www.emathzone.com/tutorials/geometry/condition-for-concurrency-of-three-straight-lines.html – lab bhattacharjee Aug 30 '14 at 17:05

score 0 · Answer 3 · answered Aug 30 '14 at 17:20

Rather than solving the equations lets try something else - since the conditions are independent of $\theta$, we can choose $\theta$ as we please.

If $\theta =0$ the first two equations give $y=1$ and $x=1$ and the third gives $l+m+n=0$. So we can have $l=m=1, n=-2$ which eliminate $a)$ and $c)$. But $b)$ becomes $-6=-6$ which is true.

Now if $\sin \theta = \cos \theta =\frac {\sqrt 2}2$ we get $x+y=2$ from the first two equations. Then if $x=0, y=2$ the third equation becomes $n=-2m$. Setting $m=n=0, l=1$ note that $b)$ becomes $1=0$, so can't always hold.

Concurrency of lines

3 Answers3