knight247 wrote:OA A
Hello!
First of all, point A definitely needs to be in Quadrant II or Quadrant IV. That means that the signs of the
x- and
y- coordinates can't be both positive or both negative.
(Why? Because, if the coordiates are equidistant, then the point rises above each axis at a 45-degree angle. So if both coordinates have the same sign, then point A lies in a straight line with O and B and there is no triangle OAB!)
This also means that A would be at either a 45-degree angle below the
x-axis or a 45-degree angle to the left of the
y-axis. Since B is already to a 45-degree angle to the right of the y-axis and above the
x-axis, it follows that angle AOB will be a 90-degree angle!
Therefore, the base that we can use to calculate the area of triangle OAB is OA, and the height is OB. (Or vice-versa, if you prefer.) Now, the length of OB is 4[root(2)]; this we know because it is the diagonal of a square of length 4. So, with the formula for the area of the triangle, we have:
A =
bh/2
16 =
b*4[root(2)]/2
16 =
b*2[root(2)]
8 =
b*[root(2)]
8/[root(2)] =
b
8*[root(2)]/2 =
b
4[root(2)] =
b
So, the length of OA is equal to the length of OB! This can only mean that the
x- and
y-coordinates of A also each have absolute values of 4, since the angle in question is 45 degrees. The possible coordinates of A in quadrants II and IV are accordingly respectively (-4, 4) and (4, -4).
Hope this helps!
Academic Coordinator
MBA Center Paris
