<p>Wow!</p>
<p>I have absolutely no idea how to do this. I got it from Question of the day:</p>
<p>Four distinct lines lie in a plane, and exactly two of them are parallel. Which of the following could be the number of points where at least two of the lines intersect?</p>
<p>Three
Four
Five</p>
<p>And the answer was three and five!~!~!</p>
<p>any help appreciated</p>
<p>thank you</p>
<p>So, you have 4 distinct lines and at least two of them are parallel. So if there are lines a,b,c,d let a||b. We know that c is not parallel to a or b (the two situations are equivalent because a||b) then c intersect both a and b (they are parallel). So so far you have two points of intersection. Let a intersects c = A and b intersects c = B. Let's think about d. If d intersects a and b in points of intersection - C and D such as C != A and D != B so you have 4 points of intersection. But we also have that ONLY two lines are parallel so c and d are intercepted. So on the whole you have 5 points of intersection. The second case is when A = C for example => then c intercepts d = A so you have only 3 points of intersection. About the other answer (4) it can't happen because if there are 4 points of intersection then c and d should be parallel and it's a contradiction with the problem because it's said that there are ONLY two parallel lines.</p>
<p>thank you very much</p>
<p>i think i get it now</p>
<p>but i still dont get why c has to intersect a and b</p>
<p>sorry if its a dumb question</p>
<p>For two lines not to intersect each other they must be parallel.
So if c intersects a then that means c is not parallel to a which again means that c is not parallel to b either.So if c is not parallel to b then the two lines will have to intersect each other at some point. Hence c has to intersect a and b.</p>
<p>it would be much easier to solve this question if you actually draw a diagram.</p>
<p>That question perfectly illustrates why the SAT I is considered a test of an applicant's reasoning ability.</p>
<p>SAT math has almost nothing to do with math.THe fact that CB gives u all formulas u need talks alot...</p>
<p>for that question, drawing a diagram is the key.
yeah a lot of SAT Math questions are like that....urgh...</p>
<p>
[quote]
That question perfectly illustrates why the SAT I is considered a test of an applicant's reasoning ability.
[/quote]
</p>
<p>It also illustrates why the SAT is problematic for somewhat intelligent hard-workers who will probably be robbed of an opportunity to attend a highly reputable college/university. With questions like this, there is only so much time to think, and it's usually never enough.</p>
<p>i thought it was stupid because it didnt say that the two other lines werent parrallel. It just said two of them were. So i thought it was kind of a trick question and if you can get another pair of pararellel lines you can make intersections in three, four, and five places.</p>
<p>They key word is "exactly"</p>
<p>If you miss that word, you will pick the wrong answer, assuming that you make the other 2 lines parallel as one of the scenarios. Read carefully!</p>
<p>the key word, as Bigb14 said, is that you have EXACTLY two parallel lines (a and b) so c has to intersect a hence b (a||b)</p>
<p>wow thanks for the help</p>
<p>personally, i think this is a stupid question. Ughhhh</p>
<p>so just to confirm i get how u have 5 points, but u can also have three, and that is if C is equal to A?</p>
<p>i dont get what u mean by that</p>
<p>You can have three points when the two unparallel lines intersect with one of the parallel lines at the same point</p>