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sujayb
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AK
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from (1)
(3r+2 -s)(4r +9 -s)= 0
on expanding it, we get

12*r^2 + s^2 - 7rs +35r - 11s + 18 = 0 ..... (a)

similarly expanding (2), we get

12*r^2 + s^2 - 7rs - 10 r +4s -12 = 0 ......(b)


now a - b will give

45*r - 15*s + 30 = 0

0r, s= 3r +2


given equation of line y= 3x+2

so we can say (r,s) lies on the line
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sujayb
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AK
from (1)
(3r+2 -s)(4r +9 -s)= 0
on expanding it, we get

12*r^2 + s^2 - 7rs +35r - 11s + 18 = 0 ..... (a)

similarly expanding (2), we get

12*r^2 + s^2 - 7rs - 10 r +4s -12 = 0 ......(b)


now a - b will give

45*r - 15*s + 30 = 0

0r, s= 3r +2


given equation of line y= 3x+2

so we can say (r,s) lies on the line


Thanks, AK. Does my method above seem plausible?
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AK
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sujayb
AK
C

Both statement alone is not sufficient
but on subtracting (2) from (1) we get s= 3r +2 which proves (r,s) lies on y=3x +2

Can you elaborate a little bit more?

From (1), I get s = 3r + 2 or s = 4r + 9
From (2), I get s = 4r - 6 or s = 3r + 2

When we combine 1 & 2 , we get s = 3r + 2. Is this what you meant ?


Your method is convincing ... this is the fastest way to solve



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