IMO D

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

Therefore both statements alone are SUFF

Wots the OA btw

I also chose D, but the OA is : C

mals24 wrote: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)

Answer C