gaurav2k101 wrote:
In the xy-plane, does the line with equation y = 3x + 2 contains the point (r,s)?
(1) \((3r+2-s)(4r+9-s)=0\)
(2) \((4r-6-s)(3r+2-s)=0\)
(r,s) is a point on the XY plane e.g (1, 2) or (2, 8) or (3, 21) etc. There is a defined relation between r and s.
Say if (r, s) is (1, 2), we can say that r + s = 3, s - r = 1, 2r + s = 4 etc. Each of these expressions will take only one value because both r and s have defined values.
This is an important point to understand in this question.
We want to know whether (r, s) lies on y = 3x + 2 i.e. whether s = 3r + 2. So we want to know whether the value of 3r - s is -2.
Note that all points (r, s) that satisfy this condition (3r - s = -2) will lie on the line y = 3x + 2. So the line represents the locus of all such points. e.g. (1, 5), (2, 8), (3, 11) etc.
Coming back to the question, we want to figure out whether 3r - s is -2.
(1) \((3r+2-s)(4r+9-s)=0\)This tells us that either 3r - s = -2 or 4r-s = -9. At least one of these relations certainly holds.
Not sufficient alone.
(2) \((4r-6-s)(3r+2-s)=0\)This tells us that either 4r - s = 6 or 3r - s = -2. At least one of these relations certainly holds.
Not sufficient alone.
Using both statements, now we know that one of 3r - s = -2 and 4r-s = -9 certainly holds and one of 4r - s = 6 and 3r - s = -2 certainly holds.
Now there are two possibilities.
Either 3r - s = -2 in which case, all works out.
Or 3r - s is not equal to -2 and both 4r-s = -9 and 4r - s = 6 hold. But here is the problem: both 4r-s = -9 and 4r - s = 6 cannot hold. As discussed above, 4r-s can take only one value. It cannot be both -9 and 6. Hence this case is not possible.
Then 3r - s must be equal to -2 and hence (r, s) must lie on the line y = 3x + 2.
Answer (C)
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