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  • A straight line L through the origin meets the lines <img alt="\mathrm{x+y=1 \; and \; x+y=3 }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx&plus;y%3D1%20%5C%3B%20and

A straight line L through the origin meets the lines \mathrm{x+y=1 \; and \; x+y=3 } at P and Q respectively. Through P and Q two straight lines \mathrm{L_1 } and \mathrm{L_2 } are drawn, parallel to \mathrm{2 x-y=5 \; and\; 3 x+y=5} respectively. Lines \mathrm{L_1 \; and\: L_2} intersect at R. Then the locus of R, as L varies, is a:

Option: 1

Parabola


Option: 2

Straight line 


Option: 3

Circle 


Option: 4

Ellipse


Answers (1)

best_answer

The line y = mx meets the given lines in P \mathrm{\left(\frac{1}{m+1}, \frac{m}{m+1}\right) \text { and } Q\left(\frac{3}{m+1}, \frac{3 m}{m+1}\right)}

Hence equation of \mathrm{L_{1}} is \mathrm{y-\frac{m}{m+1}=2\left(x-\frac{1}{m+1}\right) \Rightarrow y-2 x-1=\frac{-3}{m+1}}\; \; \; \; \; \; \; \; \; \; \; ....(1)

and that of \mathrm{L_{2}} is \mathrm{y-\frac{3 m}{m+1}=-3\left(x-\frac{3}{m+1}\right) \Rightarrow y+3 x-3=\frac{6}{m+1}}\; \; \; \; \; \; \; \; \; \; \; \; \; .....(2)

From (1) and (2) \mathrm{\frac{y-2 x-1}{y+3 x-3}=-\frac{1}{2}}

\mathrm{\Rightarrow x-3 y+5=0} which is a straight line.

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mansi

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