过AB两点的直线方程是
y+1
3+1 =
x-4
-2-4 .
点斜式为:y+1=-
2
3 (x-4)
斜截式为:y=-
2
3 x+
5
3
截距式为:
x
5
2 +
y
5
3 =1.
故答案为:
y+1
3+1 =
x-4
-2-4 ;y+1=-
2
3 (x-4);y=-
2
3 x+
5
3 ;
x
5
2 +
y
5
3 =1.
过AB两点的直线方程是
y+1
3+1 =
x-4
-2-4 .
点斜式为:y+1=-
2
3 (x-4)
斜截式为:y=-
2
3 x+
5
3
截距式为:
x
5
2 +
y
5
3 =1.
故答案为:
y+1
3+1 =
x-4
-2-4 ;y+1=-
2
3 (x-4);y=-
2
3 x+
5
3 ;
x
5
2 +
y
5
3 =1.