Question

如图已知平面直角坐标系内的三个点,A(-3,2),B(-1,3),O(0,0)是原点,求角AOB面积

Answer 1

AB的方程为(y - 2)/(x + 3) = (3 - 2)/(-1 + 3) = 1/2

y = (x+3)/2 + 2

与y轴的交点为C(0,7/2)

与x轴的交点为D(-7,0)

△CDO的面积S1 = (1/2)|OC|*|OD| = (1/2)*7*7/2 = 49/4

△ADO的面积S2 = (1/2)|OD|*OD上的高= (1/2)*7*|A的纵坐标| = （7/2)*2 = 7

△CBO的面积S3 = (1/2)|OC|*OC上的高= (1/2)*(7/2)*|B的横坐标| = （7/4)*|-1|= 7/4

△ABO的面积S = S1 - S2 -S3 = 49/4 - 7 - 7/4 = 7/2

也可以算AB的长(√5)和AB的方程,然后求O与AB的距离(7/√5).