求两条平行线5x+2y-5=0和10x+4y+35=0间的距离

L1: 5x+2y-5=0
when x =1 , y = 0
(1,0) is on L1
slope of L1= -5/2
slope of line ⊥ L1 = 2/5
line L3 through (1,0) with slope 2/5
y = (2/5)x +c
0 = 2/5 +c
c = -2/5
L3: y = (2/5)x - 2/5
solve
y = (2/5)x -2/5 and 10x+4y+35 =0
=> 10x+(4/5)x - 4/5 +35 =0
(54/5)x - 171/5 =0
x = 171/54 = 19/6
y = (2/5)(19/6) - 2/5 = 19/15 - 2/5 = 13/5
shortest distance = distance between (1,0) and ( 19/6 , 13/5)
= √ (225/36 + 169/25)
= √ 214209 /30

将后一条直线方程化为5x+2y+35/2=0
根据平行直线距离公式
d=|C1-C2|/√A^2+B^2=(45/2)/√29=45/(2√29)