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

let the eq of line L be y =mx , here m is the slope  of line..                  (m is variable)

 point of intersection of L & x+y=1 is P & point of intersection of L & x+y=3 is Q .....

               then  P =( 1/m+1  , 1/m+1 )  &

                       Q= ( 1/m+3  , 1/m+3)

line through P is parallel to 2x-y=5 so its slope is same as that of line....

          m_p=2

 line through Q is parallel to 3x+y=5 so its slope is same as that of line.....

         m_q = -3

eq of line passing through P is L_{1 = (y-(1/m+1) )} = (x-(1/m+1)).2     ........................1

eq of line passing through Q is L₂ = (y-(1/m+3)) = (x-(1/m+3)).(-3)                ...........................2

point of intersection of these lines is  R (X,Y)

          then after solving 

        X = 13/5(m+1)        and Y=21/5(m+1)

locus of R is          Y=21X/13          which is a straight line passing through origin

this answer is wrong. correct answer is x-3y+5=0

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