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Aman Singh Grade: 11

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.

7 years ago

Answers : (3)

vikas askiitian expert
510 Points

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....


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

         mq = -3

eq of line passing through P is L1 = (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



7 years ago
8 Points

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


4 years ago
Shubham Sinha
24 Points
										Vikas chutiya hai bahut chutiya hai chutiya chutiya hai  hai chutiya hai chutiya hai chutiya hai chutiya hai chutiya hai chutiya hai
one year ago
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