Question

求这道高中数学题

Answer 1

(1)若P0(1,0)，直线l:−4x+3y−1=0；

则点P0到直线l的距离d=|−4−1|(−4)2+32−−−−−−−−√=55=1.

(2)若P0(−2,1)，直线l:2x−3y=0；

则点P0到直线l的距离d=|−2×2−3|22+(−3)2−−−−−−−−√=713−−√=713−−√13

(3)若P0(2,−3),直线l:y=12x−32，即x−2y−3=0.

则点P0到直线l的距离d=|2−2×(−3)−3|1+(−2)2−−−−−−−−√=55√=5√

Answer 2

(1)
P0=(1,0)
d = | [-4(1) + 3(0) -1 ]/√(3^2+4^2) | = 1
(2)
P0=(-2,1)
d = | [2(-2) - 3(1) ]/√(2^2+3^2) | = 7/√13 =7√13/13
(3)
P0=(2,-3)
d = | [(1/2)(2)+3- 3/2 ]/√[(1/2)^2+1^2] | = √5

Answer 3

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