Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then (QR)2 is equal to .
Answer
28
Solution
Take the lines y=0, y=5, P(0,1), Q(q,0), R(p,5)
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PQ2=PR2 q2=p2+15; PQ2=QR2 2pq=p2+24
Eliminating q: p4+4p2-192=0 p2=12 q2=27
(QR)2=PQ2=q2+1=28
Answer: 28