Equation of the line : y = 3x + 2
Statement 1: (3r + 2 - s)(4r + 9 - s) = 0
Implies either,
(1) s = 3r + 2 => (r, s) lies on y = 3x + 2
or
(2) s = 4r + 9 => (r,s) does not lies on y = 3x + 2
Not sufficient.
Statement 1: (4r - 6 - s)(3r + 2 - s) = 0
Implies either,
(1) s = 4r - 6 => (r, s) does not lies on y = 3x + 2
or
(2) s = 3r + 2 => (r,s) lies on y = 3x + 2
Not sufficient.
1 & 2 Together: We get, s = 3r + 2
Therefore, (r,s) lies on y = 3x + 2
The correct answer is C.
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Hi,
For the point (r,s) to belong to the line: y = 3x+2 then
s = 3r + 2, or
3r + 2 - s = 0
1. As (3r + 2 - s )( 4r + 9 - s) = 0, then
either (3r + 2 - s) or (4r + 9 - s) can be equal to Zero. So we cannot define whether (3r + 2 - s) is equal to Zero.
Hence, 1 is insuff.
2. (4r - 6 - s) (3r + 2 - s) = 0
The same as 1, either (3r + 2 - s) or (4r - 6 - s) can be equal to Zero
Hence, 2 is insuff.
1&2
either (3r + 2 - s) or (4r + 9 - s) can be equal to Zero
AND
either (3r + 2 - s) or (4r - 6 - s) can be equal to Zero
Then 3r + 2 - s must be equal to Zero. Why?
As both (4r + 9 - s) and (4r - 6 - s) cannot be equal to Zero, or - say in another way:
4r - s cannot be both -9 and 6.
Then point (r,s) belong to the above line. Thus 1&2 is suff.
Pick C
For the point (r,s) to belong to the line: y = 3x+2 then
s = 3r + 2, or
3r + 2 - s = 0
1. As (3r + 2 - s )( 4r + 9 - s) = 0, then
either (3r + 2 - s) or (4r + 9 - s) can be equal to Zero. So we cannot define whether (3r + 2 - s) is equal to Zero.
Hence, 1 is insuff.
2. (4r - 6 - s) (3r + 2 - s) = 0
The same as 1, either (3r + 2 - s) or (4r - 6 - s) can be equal to Zero
Hence, 2 is insuff.
1&2
either (3r + 2 - s) or (4r + 9 - s) can be equal to Zero
AND
either (3r + 2 - s) or (4r - 6 - s) can be equal to Zero
Then 3r + 2 - s must be equal to Zero. Why?
As both (4r + 9 - s) and (4r - 6 - s) cannot be equal to Zero, or - say in another way:
4r - s cannot be both -9 and 6.
Then point (r,s) belong to the above line. Thus 1&2 is suff.
Pick C
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
