If you want to check if the point (r,s) lies in the line y=3x+2 substitute r in place of x and s in place of y.
Hence your equation will be s = 3r+2
Now see which of the options give you this equation
St 1: (4r+9-s)(3r+2-s) = 0
divide both sides with (4r+9-s) you get: 3r+2-s = 0
which is the same as 3r+2 = s ---> this is the equ we have to get hence SUFF
St 2: do the same thing for this st as well and you will get the equ 3r+2 = s
Suff
IMO D
St 1: (4r+9-s)(3r+2-s) = 0
divide both sides with (4r+9-s) you get: 3r+2-s = 0
which is the same as 3r+2 = s ---> this is the equ we have to get hence SUFF
I think this is main mistake
From the main statement we should find the function that will look like
s = 3r+2
(1)
(4r+9-s)(3r+2-s) = 0
means that both (4r+9-s) and (3r+2-s) can be equal to zero.
You should not divide by (4r+9-s)
(4r+9-s)(3r+2-s) = 0
4r+9-s = 0 we get s = 4r+9 this function do not contain point (r;s)
3r+2-s = 0 we get s = 3r+2 this function contain point (r;s)
INSUFF
(2)
(4r-6-s)(3r+2-s) = 0
4r-6-s = 0 we get s = 4r-6 this function do not contain point (r;s)
3r+2-s = 0 we get s = 3r+2 this function contain point (r;s)
INSUFF
(1) and (2)
we have
3r+2-s = 0 we get s = 3r+2 this function contain point (r;s)
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