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Ask community Community Discussion Question: A straight line L through the origin meets the lines x+y=1 and x+y=3 at P and Q respective
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Sharan (0)

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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    and   are drawn, parallel to 2x-y=5 and 3x+y=5 respectively. Lines    and   intersect at R. Show that the locus of R, as L varies, is a straight line.

    
akki ~~ unlucky forever ~~ (1635)

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let L :  y = mx

 

so,   P(\frac{1}{m+1},\frac{m}{m+1})\ and\  Q(\frac{3}{m+1},\frac{3m}{m+1})

 

let point R be R(h,k)

 

so by given cond.

 

2h - k =( \frac{2-m}{m+1})  and 

 

3h + k =( \frac{9+3m}{m+1})

 

eleminate 'm' from both equn & u will have locus of R . . .

 


all's well that end's well, but if it dosen't, then it is not the end . . .
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akki ~~ unlucky forever ~~ (1635)

Olaaa!! Perrrfect answer. 275  [405 rates]

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total posts: 491    
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let L :  y = mx

 

so,   P(\frac{1}{m+1},\frac{m}{m+1})\ and\  Q(\frac{3}{m+1},\frac{3m}{m+1})

 

let point R be R(h,k)

 

so by given cond.

 

2h - k =( \frac{2-m}{m+1})  and 

 

3h + k =( \frac{9+3m}{m+1})

 

eleminate 'm' from both equn & u will have locus of R . . .

 


all's well that end's well, but if it dosen't, then it is not the end . . .
  this reply:   2 points  (with Olaaa!! Perrrfect answer.   in 1   votes   )     [?]
 
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