Bunuel wrote:
If \((x_1, y_1)\), \((x_2, y_2)\) and \((x_3, y_3)\) are three vertices of a triangle ABC, what is the area of the triangle ABC?
(1) \(x_1 = -2\), \(x_2 = -3\), \(x_3 = 4\), \(y_1 = 3\), \(y_2=-1\), \(y_3=-5\)
(2) \(x_1*y_2=2\), \(x_2*y_3=15\), \(x_3*y_1=12\), \(x_3*y_2=-4\), \(x_1*y_3=10\) and \(x_2*y_1=-9\)
(1) we have the coordinates for all three vertices. this is enough info to calculate the area. we can use the distance formula to calculate side lengths of our triangle, and very easily construct a rectangle around our triangle. Find the area of the rectangle, and then subtract the area of the three associated right triangles that are formed. Sufficient
(2) We have 6 equations here, for 6 variables, but these are all proportional. we have enough info to reduce this system down in terms of one of the variables. For example, we can reduce this system down in terms of y3, for example x1 = 10/y3, x2 = 15/y3, x3 = -20/y3 y1 = -3y3/5, y2 = y3/5, and y3=y3
so our coordinates would be (10/y3, -3y3/5), (15/y3, y3/5), and (-20/y3,y3). All of these depend on variable y3, so depending on our choice of y3, we will have different coordinates and a different area. You could reduce the system down in terms of any of the other 5 variables. the main point here, is that we will never be able to know the coordinates, because our coordinates will be in terms of one of our variables, so we will get different areas depending on our choice of that variable. NS
OA is A