Q. The distance between two parallel lines is unity.A point "P" lies between the two lines at a distance "A" from one of them.Q & R are points selected , 1 on each line such that triangle PQR is equilateral and the side of the equilateral triangle is 2√7/3 then A=
Please explain how you do it and maybe even show the diagram too.
62assume that point Q is the origin
and the line with point Q is the X axis..
then point R has a Y coordinate of 1
Now try to solve it :)
Edited after the next post...
62sorry a mistake
R has y corrdiate of 1
y coordiante of P is between 0 and 1
3But then if P , Q & R in the same line , then how can a triangle be formed ?
62tormented..
see carfully only the y coordinates are like that
the x coordinates take different or intermediate values...
1Use trignometry for this
drop prependiculars from P to both the lines..let it be D1 and D2
Q on line 1, R 0n 2 ..distance of P from 1 = p , from 2 = 1-p
Let angle QPD1 = angle(fi) , angle RPD1 = angle(a;[ha)
see that angle( fi + alpha) = 120
= fi = 120-alhpa ---------------------(1)
also let side of equilateral triangle be "a"
then acos(alpha) = 1-p
ascos(fi) = p ( can get sin(fi) from here only)
now it can be solved..take cos on both sides ..
1see that angle( fi + alpha) = 120
becos fi + angleQPR + alpha = 180
angle QPR = 60 degree
hence the statement..
its difficult for me to draw the figure here.. :P
3Hmm yes but is there any of proving it using distance formula or section formula or any of the basic coordinate fourmulae ?
62All of maths is interlinked...
So you should be smart to be able to shift topics when solving problems..
like problems of Coordinate using trigo or complex or even vectors...
Just dive deep into the sea called Mathematics and you will love it. If you keep hitting the surface you will be blown by the waves ;)