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高中数学平面解析几何初步2.1_2.1.5平面上两点间的距离练习苏教版

2.1.5 平面上两点间的距离

A 组 基础巩固

1.已知 A(-1,0),B(5, 6),C(3,4),则||ACCB||的值 为(

)

A.13

B.12

C.3

D.2

解析:由两点间的距离公式,得

|AC|= [3-(-1)2]+(4-0)2=4 2,

|CB|= (3-5)2+(4-6)2=2 2,

故||ACCB||=24

2=2. 2

答案:D 2.直线 y=x 上的两点 P,Q 的横坐标分别是 1,5,则|PQ|等于( )

A.4 B.4 2 C.2 D.2 2

解析:由题意易知 P(1,1),Q(5, 5),

所以|PQ|= 2(5-1)2=4 2.

答案:B

3.已知点 A(-2,-1),B(a,3)且|AB|=5,则 a 等于( )

A.1 B.-5 C.1 或-5 D.其他值 解析:由两点间距离公式得,(a+2)2+(3+1)2=52, 所以(a+2)2=9.所以 a=1 或 a=-5.

答案:C

4.与两点 A(-2,2),B(2,4)等距离,且在坐标轴上的点的坐标是________________.

解析:设点 P(a,0)或(0,b)由两点间的距离公式计算.

答案:???32,0???和(0,3) 5.光线从点 A(-3,5)射到直线 l:3x-4y+4=0 以后,再反射到一点 B(2,15).

(1)求入射线与反 射线的方程; (2)求这条光线从 A 到 B 的长度. 解:(1)设点 A 关于直线 l 的对称点为 A1(x0,y0),由直线 AA1 与已知直线垂直,且 AA1 中点也在直线上,则有

??yx00-+53=-43, ???3×x0-2 3-4×y0+2 5+4=0,

解得 x0=3,y0=-3, 即 A1(3,-3).
y+3 x-3 于是反射光线方程为15+3=2-3,

即 18x+y-51=0. 同理 B1(14,-1),入射光线方程为 6x+17y-67=0. (2)光线从 A 到 B 的长度,利用线段的垂直平分线性质,即得 AP+PB=A1P+PB=A1B= (3-2)2+(-3-15)2=5 13.

B 级 能力提升 6.x 轴上任一点到定点(0,2),(1,1)距离之和的最小值是( )

A. 2 B.2+ 2 C. 10 D. 5+1

解析:作点(1,1)关于 x 轴的对称点(1,-1),则距离之和最小值为 12+(-1-2)2

= 10.

答 案:C

7.直线 y=3x-4 关于点 P(2,-1)对称的直线 l 的方程是( )

A.y=3x-10

B.y=3x-18

C.y=3x+4

D.y=4x+3

解析:在直线上任取两点 A (1,-1),B(0,-4),则其关于点 P 的对称点 A′,B′可

由中点坐标公式求得为 A′(3,-1),B′(4,2).由两点式可求得方程为 y=3x-10.

答案:A 8.已知 A(-3,8),B(2,2),在 x 轴上有一点 M,使得|MA|+|MB|最短,则点 M 的坐

标是( )

A.(-1,0)

B.(1,0)

C.???252,0???

D.???0,252???

解析:如图所示,A 关于 x 轴的对称点为 A′(-3,-8),则 A′B 与 x 轴的交点即为 M,

求得 M 的坐标为(1,0).

答案:B

9.已知 A(5,2a-1)、B(a+1,a-4),则 AB 的最小值是________.

解析:由两点间的距离公式得

AB= (a-4)2+(a-4-2a+1)2=

2a2-2a+25=

2???a-21???2+429≥7 2 2.

答案:7 2 2

10.点 A(-3,1),C(1,y)关于点 B(-1,-3)对称,则 AC=______.

y+1 解析:由已知得 2 =-3,所以

y=-7,即

C(1,-7),

所以|AC|= [1-(-3)2]+(-7-1)2=4 5.

答案:4 5 11.直线 x-2y+1=0 关于直线 y-x =1 对称的直线方程是____. 解析:设所求直线上任一点的坐标为(x,y),则它关于 y-x=1 对称的点为(y-1,x +1),且在直线 x-2y+1=0 上, 所以 y-1-2(x+1)+1=0,化简得 2x-y+2=0. 答案: 2x-y+2=0

1 2.甲船在某港口东 50 km,乙船在同一港口的东 14 km,南 18 km 处,那么甲、乙两

船的距离是________.

解析:以某港口为坐标原点建系后得甲船坐标为(50,0),乙船坐标为(14,-18),由

两点间距离公式 得甲、乙两船的距离为 (50-14)2+(0+18)2=18 5.

答案:18 5 km 13.已知点 A(a+1, 2)与点 B(5,a)的距离为 2,求 a 的值.

解:由已知得|AB|= [5-(a+1)]2+(a-2)2=2 , 即(a-4)2+(a-2)2=4,化简整理得 a2-6a+8=0, 解之得 a=2 或 a=4. 14.已知 0<x<1,0<y<1.

求证: x2+y2+ x2+(1-y)2+ (1-x)2+y2+ (1-x)2+(1-y)2≥2 2,

并求等号成立的条件. 证明:设四边形 OABC 是正方形,O(0,0),A(1,0),B(1,1),C(0,1),设 P(x,y)

为正方形内一点,如图所示,

|PO|= x2+y2,|PA|= (1-x)2+y2,

|PB|= (1-x)2+(1-y)2,PC= x2+(1-y)2,

OB= 2,AC= 2.

因为 PO+PB≥BO,PA+PC≥AC, 所以 PO+PB+PA+PC≥BO+AC, 即 x2+y2+ x2+(1-y)2+ (1-x)2+y2+
(1-x)2+(1-y)2≥2 2.当且仅当 PO+PB=BO, 且 PA+PC=AC 时,等号成立.此时点 P 既在 OB 上, 又在 AC 上,因此,点 P 是正方形的中心,即 x=y=12.


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