A（-3,40）在一次函数y=-3x-5的图像中,与y轴交点为b,求三角形AOB的面积

y = -3x- 5
y(0) = -5
B(0,-5)
|AB| = √1609
let H(x,y) be a point on AB,then
(y-40)/(x+3) = (-5-40)/(0+3)
y-40 = -15(x+3)
y = -15x - 5
then H(x,-15x-5)
|HO|^2 = x^2 + (-15x-5)^2
(|HO|^2)' = 2x +30(15x+5) = 0 ( if OH is perpendicular to AB)
x+225x+ 75 = 0
x = -75/226
y = -15( -75/226) -5 = -5/226
H(-75/226,-5/226)
|OH| = √[(-75/226)^2+(-5/226)^2]
=5 (√226)/226
三角形AOB的面积
=(1/2) OH .AB
= (1/2)5 (√226)/226.√1609
=√363634/2260