georgethomps wrote:
m1033512 wrote:
You may have combination of 1,4 also which will satisfy the given equation .
BUT THIS IS NOT CORRECT
Actually given equation is a collection of 4 straight lines which are caluclated on below logic
when x and y both negative
x negative and y positive
x positive and y negative
x positive and y positive
this way we get four lines amd then draw them on the xy plane
and calculate the enclosed area .
award kudos if helpful
Posted from my mobile device
I'm still confused. Possible combinations of (5,0), (4,1), (3,2) all yield inconsistent results, wouldn't this be an invalid problem? What is wrong about using (1,4) or (2,3)? They both satisfy the equation and yield straight lines on the graph. I'm guessing this is supposed to be wrong because only (0,5) doesn't plot vertical lines, but I don't see anything in the problem that prevents this. The problem seems to state plotting the coordinates and interpreting them visually.
Note that the question has asked for the
area enclosed by the equation. In other words it is asking for the area where all these lines (4 in this case) intersect. From the equation we can see that the lines will intersect at (5,0), (0,5), (-5,0) and (0,-5).
While your coordinates may satisfy the equation, they are not the points at which these lines intersect. To make it simpler, when asked to compute area under the graph in questions of this type, look for the coordinates at which the lines intersect and compute the area enclosed by them.
Hope its clear now.