Question

已知反比例函数图像与直线AB相交于A、B两点,其中A点的坐标为(1,4),连接OA、OB,当三角形AOB的面积为15/2

Answer 1

y=k/x

A(1,4)

k=1*4=4

y=4/x

y-4=k'(x-1)

[k'(x-1)+4]x=4

k'x^2+(4-k')x-4=0

(k'x+4)(x-1)=0

x= -4/k' y=-k'

B(-4/k',k')

Soab=(-k'+4)(-4/k'-1)/2+1*4/2-(-4/k')(-k')=15/2

k'-16/k'=15

k'^2-15k'-16=0

(k'+1)(k'-16)=0

k'=-1

直线为：y-4=-1(x-1)

y=-x+5