写出满足下列条件的直线的方程(1)斜率是3分之√3,经过点A(8,-2)(2)见过点B(-2,0),且与x轴垂直(3)斜
2个回答

(1)

(y-y1)=k(x-x1)

y+2=√ 3/3(x-8)

y=(√ 3/3)x-(8√3/3)-2

(2)

x=-2

(3)

y=kx+b

k=斜率,b=y轴截距

y=-4x+7

(4)

(y-y1)/(y2-y1)=(x-x1)/(x2-x1)

y-8/(-2-8)=(x-(-1)/(4-(-1)))

y=-2x+6

(5)

y=2

(6)

当直线不过原点时 x/a+y/b=1

当直线过原点时 y=kx

x/4-y/3=1