题目

已知点A(1,y1),B(2,y2),C（3,y3）都在抛物线y=5x²+12x上,则△ABC的面积是____.

提问时间：2020-10-10

答案

y1 = 5 *1² + 12 * 1 = 17
y2 = 5 *2² + 12 * 2 = 44
y3 = 5 *3² + 12 * 3 = 81
BC直线 k = (81-44)/(3-2) = 37
y-81 = 　37 (x - 3)
BC直线交y=17的直线,交点　P
y-81 = 　37 (x - 3)　
y = 17
得x = 3-64/37
交点　P(3-64/37,　17)
BP = 3-64/37 -1 = 2-64/37
ABC的面积 = APC的面积 -BPC的面积 = 1/2 * BP *(81-17) - 1/2 * BP * (44-17)