Problem: A line through the point ( −a,0 ) cuts from the second quadrant a triangular region with area T. The equation of the line is:
Answer Choices:
A. 2Tx+a2y+2aT=0
B. 2Tx−a2y+2aT=0
C. 2Tx+a2y−2aT=0
D. 2Tx−a2y−2a T=0
E. none of these
Solution:
Let the line cut the x-axis in (−a,0) and the y-axis in (0,h). Then 21ah=T and h=2T/a. The slope of the line is ah=a2T. Therefore, y=a22Tx+a2T∴2Tx−a2y+2aT=0