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Question

A straight line L through the origin meets the line x+y=1 and x+y=3 at P and Q respectively.Through P and Q two straight lines L1 and L2 aredrawn, parallel to 2xy=5 and 3x+y=5respectively. Lines L1 and L2 intersect at R. Show that the locus of R, as L varies, is a straight line.

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Solution

Any line through origin is y=mx (m variable). It meets the given lines in points.
P(1m+1,mm+1) and Q(3m+1,3mm+1)

Any line L1 through P parallel to 2xy=5 is
2(x1m+1)(ymm+1)=0

Or 2xy=2mm+1

Any line through Q parallel to 3x+y=5 is
3(x3m+1)+(y3mm+1)=0

Or 3x+y=3(3+m)m+1

In order to find the locus of their point of intersection, we have to eliminate the variable m between their equations. We rewrite the equations of lines as
2xy=m1+3m+1 or 2xy+1=3m+1

3x+y=3(2+m+1)m+1 or 3xy3=6m+1

Dividing the above two thereby eliminating the variable m, the required locus is given by

2xy+13x+y3=12

Or 4x2y+2=3x+y3

Or x3y+5=0

Which being of first degree, represents the equation of a line.


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