Question

1.写出满足下列条件的直线的点斜式方程：

Answer 1

⑴(y+1)=4*(x-3)

⑵(y+2)=-2*(x-2)

⑶tanα=-√3=k ,(y-2)=-√3*(x+4)

⑷α=0°,∴斜率k=0,(y+1)=0*(x-3)

⑴y=-2x

⑵∵平行与X轴,所以斜率k=0,(y-2)=0*(x+6)

⑶∵平行于Y轴,∴斜率不存在,x=2

⑴(x-0)/((0+3)=(y-0)/(0-1)

⑵(x+6)/(-6+4)=(y-2)/(y+2)

⑶(x-2)/(2+4)=(y-2)/(2-2)

⑴(y-4)=-2*(x-0)

⑵在Y轴上截距=-4,则该直线交Y轴于（0,-4）,(x-2)/(1-0)=(y+1)/(-1+4)

⑶(y-3)=-1(x-0)

⑷(x-5)/(5-0)=(y-0)/(0-4)