dpatwa wrote:In the xy-plane, does the line with equation y=3x+2 contain the point (r,s)?

1. (3r+2-s)(4r+9-s)=0

2. (4r-6-s)(3r+2-s)=0

[spoiler]OA: C[/spoiler]

Is there a trick to this? Is multiplying the expression out necessary?

This question has been answered before .. many times .. anyway here goes ...

For any line with the equation y = 3x+2, we can say that the point ( r, s) lies on it if r and s satisfy the equation .. so for example r = 1 and s = 5, then the equation will become 5 = 3*1 +2 , that is the equation is satisfied so the point (1,5) does lie on the line y = 3x+2 , but the point (2,5) will not line because it does not satisfy the equation .. that was the basics ..

Now let us take a look at the question ..

the first statement says that (3r+2-s)(4r+9-s)=0 .. that means either (3r+2-s ) = 0 or (4r+9-s) = 0 , if 3r+2-s = 0 then the point (r,s) does lie on the line because as can be seen we get s = 3r+2 , which means (r,s ) satisfies the equation y = 3x + 2 ( Remember in the point (r,s) r represents the x coordinate and s represents the y coordinate ).. OR .. we have 4r+9-s = 0 if this is true then the point does not lie on the line as (r,s) does not satisfy the equation .. So we dont get a defnite answer ... and therefore the first statement is insufficient ..

A similar reasoning can be used for the 2nd statement ..

Combine the 2 statements and we know that in both the given equation .. it is 3r+2-s which is equal to 0 .. which means that (r,s) does lie on the line y = 3x +2 .. hence the answer is C ...