2.3.2 两点间的距离
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【基础过关系列】2022-2023学年高二数学上学期同步知识点剖析精品讲义(人教A版2019)
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选择性必修第一册同步巩固,难度2颗星!
基础知识
两点距离公式
平面上的两点\(P_1 (x_1 ,y_1)\) ,\(P_2 (x_2 ,y_2)\)间的距离公式 \(\left|P_{1} P_{2}\right|=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\).
证明 \(\left|P_{1} P_{2}\right|=\left|\overrightarrow{P_{1} P_{2}}\right|=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\).
【例】 已知点\(P(x,2)\),\(Q(-2,-3)\),\(M(1,1)\),且\(|PQ|=|PM|\),则\(x\)的值为\(\underline{\quad \quad}\).
解析 因为\(P(x,2)\),\(Q(-2,-3)\),\(M(1,1)\),且\(|PQ|=|PM|\),
所以\((a+2)^2+(2+3)^2=(a-1)^2+(2-1)^2\),
解得 \(a=-\dfrac{9}{2}\).
两点距离公式的几何意义
几何问题与代数问题间可相互转化.
形如 \(d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)的式子可以理解为点两点\(P_1 (x_1 ,y_1)\) ,\(P_2 (x_2 ,y_2)\)间的距离,
则求解与\(d\)有关的代数问题可转化为几何问题.
【例】 已知点\(M(a,b)\)在直线\(l:x+y-1=0\)上,则 \(\sqrt{a^{2}+b^{2}}\)的最小值为\(\underline{\quad \quad}\).
解析 \(\sqrt{a^{2}+b^{2}}=\sqrt{(a-0)^{2}+(b-0)^{2}}\)理解为\((a,b)\)到原点\((0,0)\)的距离,
则 \(\sqrt{a^{2}+b^{2}}\)的最小值为原点\((0,0)\)到直线\(l:x+y-1=0\)的距离
(利用后面所学点到直线距离公式可求).
基本方法
【题型1】两点间的距离
【典题1】 在直线\(2x-y=0\)上求一点\(P\),使它到点\(M(5,8)\)的距离为\(5\),并求直线\(PM\)的方程.
解析 \(∵\)点\(P\)在直线\(2x-y=0\)上,\(∴\)可设\(P(a,2a)\).
根据两点的距离公式得
\(|P M|^{2}=(a-5)^{2}+(2 a-8)^{2}=5^{2}\),即\(5a^2-42a+64=0\),
解得\(a=2\)或 \(a=\dfrac{32}{5}\),\(∴P(2,4)\)或 \(\left(\dfrac{32}{5}, \dfrac{64}{5}\right)\).
\(∴\)直线\(PM\)的方程为 \(\dfrac{y-8}{4-8}=\dfrac{x-5}{2-5}\)或 \(\dfrac{y-8}{\dfrac{64}{5}-8}=\dfrac{x-5}{\dfrac{32}{5}-5}\),
即\(4x-3y+4=0\)或\(24x-7y-64=0\).
巩固练习
1.已知点\(M(m,-1)\),\(N(5,m)\),且 \(|M N|=2 \sqrt{5}\),则实数\(m\)等于( )
A.\(1\) \(\qquad \qquad\) B.\(3\) \(\qquad \qquad\) C.\(1\)或\(3\) \(\qquad \qquad\) D.\(-1\)或\(3\)
2.已知\(△ABC\)的顶点为\(A(2,1)\),\(B(-2,3)\),\(C(0,-1)\),则\(AC\)边上的中线长为( )
A.\(3\) \(\qquad \qquad\) B. \(3 \sqrt{2}\) \(\qquad \qquad\) C.\(4\) \(\qquad \qquad\) D. \(4\sqrt{2}\)
3.已知三个点\(A(-3,1)\),\(B(3,-3)\),\(C(1,7)\),
(1)判断\(△ABC\)的形状;(2)求\(△ABC\)的面积.
参考答案
-
答案 \(C\)
解析 因为点\(M(m,-1)\),\(N(5,m)\),且 \(|M N|=2 \sqrt{5}\),
所以 \(|M N|=\sqrt{(m-5)^{2}+(m+1)^{2}}=2 \sqrt{5}\),
即\(m^2-4m+3=0\),解得\(m=1\)或\(m=3\).
故选:\(C\). -
答案 \(B\)
解析 根据题意,设\(AC\)的中点为\(D\),
\(△ABC\)的顶点为\(A(2,1)\),\(B(-2,3)\),\(C(0,-1)\),则\(D(1,0)\),
\(|B D|=\sqrt{9+9}=3 \sqrt{2}\),
故选:\(B\). -
答案 (1) 等腰直角三角形 (2) \(26\)
解析 (1) \(∵△ABC\)的三个顶点的坐标分别为\(A(-3,1)\),\(B(3,-3)\),\(C(1,7)\),
\(\therefore|A B|=\sqrt{6^{2}+4^{2}}=2 \sqrt{13}\), \(|B C|=\sqrt{2^{2}+10^{2}}=2 \sqrt{26}\), \(|A C|=\sqrt{4^{2}+6^{2}}=2 \sqrt{13}\)
\(∴BC^2=AC^2+AB^2\),\(AB=AC\),
\(∴△ABC\)是等腰直角三角形.
(2)\(△ABC\)的面积 \(S=\dfrac{1}{2} A B \cdot A C=\dfrac{1}{2} \times 2 \sqrt{13} \times 2 \sqrt{13}=26\).
【题型2】两点间距离的最值问题
【典题1】 设\(a,b∈R\), \(\sqrt{(a-1)^{2}+(b-1)^{2}}+\sqrt{(a+1)^{2}+(b+1)^{2}}\)的最小值为\(\underline{\quad \quad}\).
解析 从几何意义看, \(\sqrt{(a-1)^{2}+(b-1)^{2}}+\sqrt{(a+1)^{2}+(b+1)^{2}}\)表示点\((a,b)\)到点 \((1,1)\)和\((-1,-1)\)距离的和,其最小值为 \((1,1)\)和\((-1,-1)\)两点间的距离 \(2 \sqrt{2}\).
【典题2】已知两定点\(A(-3,5)\),\(B(2,8)\),动点\(P\)在直线\(x-y+1=0\)上,则\(|PA|+|PB|\)的最小值为( )
A.\(5 \sqrt{13}\) \(\qquad \qquad\) B.\(\sqrt{34}\)\(\qquad \qquad\) C.\(5 \sqrt{5}\)\(\qquad \qquad\) D.\(2 \sqrt{26}\)
解析 \(∵\)两定点\(A(-3,5)\),\(B(2,8)\),动点\(P\)在直线\(x-y+1=0\)上,
\(∴\)点\(A(-3,5)\),\(B(2,8)\)在直线\(x-y+1=0\)同侧,
设点\(A\)关于直线\(x-y+1=0\)的对称点为\(C(a,b)\),
则 \(\left\{\begin{array}{l}
\dfrac{a-3}{2}-\dfrac{5+b}{2}+1=0 \\
\dfrac{b-5}{a+3}=-1
\end{array}\right.\),解得\(a=4\),\(b=-2\),\(∴C(4,-2)\),
\(∴|PA|+|PB|\)的最小值为: \(|B C|=\sqrt{(4-2)^{2}+(-2-8)^{2}}=2 \sqrt{26}\).
故选:\(D\).
【典题3】在平面直角坐标系\(xOy\)中,设定点\(A(a,a)\),\(P\)是函数 \(y=\dfrac{1}{x}(x>0)\)图象上一动点.
若点\(P\),\(A\)之间的最短距离为 \(2 \sqrt{2}\),则满足条件的实数\(a\)的所有值为( )
A. \(\sqrt{10}\) \(\qquad \qquad\) B.\(\pm \sqrt{10}\) \(\qquad \qquad\) C. \(3\)或 \(-1\) \(\qquad \qquad\) D. \(\sqrt{10}\)或 \(-1\)
解析 设 \(P\left(x, \dfrac{1}{x}\right)\),
则 \(d=|P A|=\sqrt{(x-a)^{2}+\left(\dfrac{1}{x}-a\right)^{2}}\)\(=\sqrt{\left(x+\dfrac{1}{x}\right)^{2}-2 a\left(x+\dfrac{1}{x}\right)+2 a^{2}-2}\).
令 \(t=x+\dfrac{1}{x} \geq 2\)
\(\therefore d=\sqrt{t^{2}-2 a t+2 a^{2}-2}\),
令 \(f(t)=t^{2}-2 a t+2 a^{2}-2, t \geq 2\).
该函数对称轴\(t=a\)
①\(a≤2\)时,\(f(t)\)递增, \(f(t)_{\min }=f(2)=2 a^{2}-4 a+2=8\)
解得\(a=-1\)或\(3\)(舍)
②①\(a>2\)时, \(f(t)_{\min }=f(a)=a^{2}-2=8\)
解得\(a=\sqrt{10}\)或\(-\sqrt{10}\)(舍).
综上,\(a\)的取值为\(-1\)或\(\sqrt{10}\).
故选:\(D\).
巩固练习
1.函数 \(f(x)=\sqrt{x^{2}-2 x+2}+\sqrt{x^{2}-4 x+8}\)的最小值为\(\underline{\quad \quad}\).
2.已知点\(R\)在直线\(x-y+1=0\)上,\(M(1,3)\),\(N(3,-1)\),则\(∥RM|-|RN∥\)的最大值为\(\underline{\quad \quad}\) .
3.已知点\(A(1,1)\),\(B(2,2)\),点\(P\)在直线 \(y=\dfrac{1}{2} x\)上,求\(|PA|^2+|PB|^2\)取得最小值时\(P\)点的坐标\(\underline{\quad \quad}\).
4.已知点\(A(2,1)\),点\(B\)为两条直线\(y=kx-2k+1\)与\(x+y-1=0\)交点,则\(|AB|\)的最小值为\(\underline{\quad \quad}\) .
5.在平面直角坐标系\(xOy\)中,点\(A(1,0)\),\(B(4,0)\).若直线\(x-y+m=0\)上存在点\(P\)使得\(PB=2PA\),求实数\(m\)的取值范围.
参考答案
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答案 \(\sqrt{10}\)
解析 \(f(x)=\sqrt{x^{2}-2 x+2}+\sqrt{x^{2}-4 x+8}\)可看作点\((x,0)\)到
点\((1,1)\)和点\((2,2)\)的距离之和,作点\((1,1)\)关于\(x\)轴对称的点\((1,-1)\)
\(\therefore f(x)_{\min }=\sqrt{10}\). -
答案 \(\sqrt{10}\)
解析 设点\(M(1,3)\)关于直线\(x-y+1=0\)的对称点为\(M' (x,y)\),
则 \(\left\{\begin{array}{l} \dfrac{y-3}{x-1}=-1 \\ \dfrac{1+x}{2}-\dfrac{3+y}{2}+1=0 \end{array}\right.\),解得 \(\left\{\begin{array}{l} x=2 \\ y=2 \end{array}\right.\),即\(M' (2,2)\),
\(∵N(3,-1)\),
\(\therefore\|R M|-| R N\|=\left\|R M^{\prime}|-| R N\right\| \leqslant\left|M^{\prime} N\right|=\sqrt{10}\). -
答案 \(\left(\dfrac{9}{5}, \dfrac{9}{10}\right)\)
解析 设\(P(2t,t)\),
则\(|P A|^{2}+|P B|^{2}=(2 t-1)^{2}+(t-1)^{2}+(2 t-2)^{2}+(t-2)^{2}\)\(=10 t^{2}-18 t+10\)
当 \(t=\dfrac{9}{10}\)时,\(|PA|^2+|PB|^2\)取得最小值,此时有 \(P\left(\dfrac{9}{5}, \dfrac{9}{10}\right)\)
\(|PA|^2+|PB|^2\)取得最小值时\(P\)点的坐标为 \(\left(\dfrac{9}{5}, \dfrac{9}{10}\right)\). -
答案 \(\sqrt{2}\)
解析 当\(k=-1\)时,两条直线\(y=kx-2k+1\)与\(x+y-1=0\)无交点,
当\(k≠-1\)时,由\(\left\{\begin{array}{l} y=k x-2 k+1 \\ x+y-1=0 \end{array}\right.\),解得 \(\left\{\begin{array}{l} x=\dfrac{2 k}{1+k} \\ y=\dfrac{1-k}{1+k} \end{array}\right.\),
即交点坐标为 \(\left(\dfrac{2 k}{1+k}, \dfrac{1-k}{1+k}\right)\),
又\(A(2,1)\),
所以 \(|A B|=\sqrt{\left(\dfrac{2 k}{1+k}-2\right)^{2}+\left(\dfrac{1-k}{1+k}-1\right)^{2}}=2 \sqrt{\dfrac{1+k^{2}}{(1+k)^{2}}}\)\(=2 \sqrt{\dfrac{1+k^{2}+2 k-2 k}{1+k^{2}+2 k}}\)
\(=2 \sqrt{1-\dfrac{2 k}{1+k^{2}+2 k}}=2 \sqrt{1-\dfrac{2}{\dfrac{1}{k}+k+2}}\),
因为当\(k>0\)时, \(k+\dfrac{1}{k} \geqslant 2\)当且仅当\(k=1\)取等号,此时\(|AB|\)的最小值为 \(\sqrt{2}\).
当\(k<0\),且\(k≠-1\)时, \(k+\dfrac{1}{k}\)没有最小值,此时\(|AB|\)无最小值,
综上所述\(|AB|\)的最小值为 \(\sqrt{2}\). -
答案 \([-2 \sqrt{2}, 2 \sqrt{2}]\)
解析 设\(P(x,x+m)\),
\(∵2PA=PB\),\(∴4|PA|^2=|PB|^2\),
\(∴4(x-1)^2+4(x+m)^2=(x-4)^2+(x+m)^2\),
化为\((x+m)^2=4-x^2\),
\(∴4-x^2≥0\),解得\(x∈[-2,2]\),
\(\therefore m=-x \pm \sqrt{4-x^{2}}\),令 \(x=2 \cos \theta\),\(\theta \in[0, \pi]\)
\(\therefore m=-2 \cos \theta \pm 2 \sin \theta=\pm 2 \sqrt{2} \sin \left(\theta \pm \dfrac{\pi}{4}\right) \in[-2 \sqrt{2}, 2 \sqrt{2}]\),
实数m的取值范围是 \([-2 \sqrt{2}, 2 \sqrt{2}]\),
故答案为 \([-2 \sqrt{2}, 2 \sqrt{2}]\).
分层练习
【A组---基础题】
1.在直角坐标系\(xOy\)中,已知点\(A(4,2)\)和\(B(0,b)\)满足\(|BO|=|BA|\),那么\(b\)的值为( )
A.\(3\) \(\qquad \qquad\) B.\(4\) \(\qquad \qquad\) C.\(5\)\(\qquad \qquad\) D.\(6\)
2.以\(A(-1,1)\)、\(B(2,-1)\)、\(C(1,4)\)为顶点的三角形是
A.锐角三角形 \(\qquad \qquad\) B.钝角三角形 \(\qquad \qquad\)
C.以\(A\)点为直角顶点的直角三角形 \(\qquad \qquad\) D.以\(B\)点为直角顶点的直角三角形
3.直线\(l_1:x-my-2=0\)与直\(线l_2:mx+y+2=0\)交于点\(Q\),\(m\)是实数,\(O\)为坐标原点,则\(|OQ|\)的最大值是 ( )
A.\(2\) \(\qquad \qquad\) B. \(2 \sqrt{2}\) \(\qquad \qquad\) C. \(2 \sqrt{3}\) \(\qquad \qquad\) D.\(4\)
4.已知点\(A(4,12)\),点\(P\)在\(x\)轴上,且点\(A\)与点\(P\)间的距离为\(13\),则点\(P\)的坐标为\(\underline{\quad \quad}\).
5.函数 \(f(x)=\sqrt{x^{2}-4 x+5}+\sqrt{x^{2}-6 x+13}\)的最小值为\(\underline{\quad \quad}\).
6.已知点\(A(4,0)\),\(B(0,2)\),对于直线\(l:x-y+m=0\)的任意一点\(P\),都有\(|PA|^2+|PB|^2>18\),则实数\(m\)的取值范围是\(\underline{\quad \quad}\).
7.已知\(|m|<1\),直线\(l_1:y=mx+1\),\(l_2:x=-my+1\),\(l_1\)与\(l_2\)相交于点P,\(l_1\)交\(y\)轴于点\(A\),\(l_2\)交\(x\)轴于点\(B\)
(1)证明:\(l_1⊥l_2\);
(2)用\(m\)表示四边形\(OAPB\)的面积\(S\),并求出\(S\)的最大值;
参考答案
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答案 \(C\)
解析 \(∵\)点\(A(4,2)\)和\(B(0,b)\)满足\(|BO|=|BA|\),
\(∴b^2=4^2+(2-b)^2\),\(∴b=5\).
故选:\(C\). -
答案 \(C\)
解析 \(∵A(-1,1)\)、\(B(2,-1)\)、\(C(1,4)\),
\(\therefore|A B|=\sqrt{(2+1)^{2}+(-1-1)^{2}}=\sqrt{13}\), \(|A C|=\sqrt{(1+1)^{2}+(4-1)^{2}}=\sqrt{13}\),
\(|B C|=\sqrt{(1-2)^{2}+(4+1)^{2}}=\sqrt{26}\),
\(∴|AC|^2+|AB|^2=|BC|^2\),
\(∴\)以\(A(-1,1)\)、\(B(2,-1)\)、\(C(1,4)\)为顶点的三角形是以\(A\)点为直角顶点的直角三角形.
故选:\(C\). -
答案 \(B\)
解析 \(∵l_1:x-my-2=0\)与直线\(l_2:mx+y+2=0\)交于点\(Q\),
由 \(\left\{\begin{array}{l} x-m y-2=0 \\ m x+y+2=0 \end{array}\right.\),可得 \(\left\{\begin{array}{l} x=\dfrac{2-2 m}{m^{2}+1} \\ y=-\dfrac{2 m+2}{m^{2}+1} \end{array}\right.\),可得点 \(Q\left(\dfrac{2-2 m}{m^{2}+1},-\dfrac{2 m+2}{m^{2}+1}\right)\),
\(m\)是实数,\(O\)为坐标原点,
则 \(|O Q|=\sqrt{\left(\dfrac{2-2 m}{m^{2}+1}\right)^{2}+\left(-\dfrac{2 m+2}{m^{2}+1}\right)^{2}}=\sqrt{\dfrac{8\left(1+m^{2}\right)}{\left(1+m^{2}\right)^{2}}}\)\(=\dfrac{2 \sqrt{2}}{\sqrt{1+m^{2}}}\),
当\(m=0\)时, \(|O Q|_{\max }=2 \sqrt{2}\),
所以\(|OQ|\)的最大值是 \(2 \sqrt{2}\).
故选:\(B\). -
答案 \((-1,0)\)或\((9,0)\)
解析 \(∵\)点\(P\)在\(x\)轴上,\(∴\)设\(P(x,0)\),
\(∵\)点\(P\)与点\(A\)的距离等于\(13\),
\(\therefore \sqrt{(x-4)^{2}+(0-12)^{2}}=13\),解得:\(x=9\)或\(-1\),
\(∴\)点\(P\)的坐标为\((-1,0)\)或\((9,0)\). -
答案 \(\sqrt{10}\)
解析 \(f(x)=\sqrt{x^{2}-4 x+5}+\sqrt{x^{2}-6 x+13}\)可看作点\((x,0)\)到
点\((2,1)\)和点\((3,2)\)的距离之和,作点\((2,1)\)关于x轴对称的点\((2,-1)\),
\(\therefore f(x)_{\min }=\sqrt{10}\). -
答案 \((-\infty,-1-2 \sqrt{2}) \cup(-1+2 \sqrt{2},+\infty)\)
解析 根据题意,点\(P\)在直线\(l:x-y+m=0\)上,设\(P\)的坐标为\((x,x+m)\),
则有 \(|P A|^{2}+|P B|^{2}=(x-4)^{2}+(x+m-0)^{2}+(x-0)^{2}+(x+m-2)^{2}\)
\(=4x^2+(4m-12)x+(2m^2-4m+20)\),
若对于直线\(l:x-y+m=0\)上的任意一点\(P\),都有\(|PA|^2+|PB|^2>18\),则\(4x^2+(4m-12)x+(2m^2-4m+20)>18\)恒成立,
即\(4x^2+(4m-12)x+(2m^2-4m+2)>0\)对于\(R\)恒成立,
则有\(△=(4m-12)^2-16(2m^2-4m+2)<0\),
即\(m^2+2m-7>0\),
解可得 \(m>-1+2 \sqrt{2}\)或 \(m<-1-2 \sqrt{2}\),
即\(m\)的取值范围为 \((-\infty,-1-2 \sqrt{2}) \cup(-1+2 \sqrt{2},+\infty)\). -
答案 (1) 略 (2) \(1\)
解析 (1)当\(m=0\)时,直线\(l_1:y=1\),\(l_2:x=1\),显然有\(l_1⊥l_2\);
当\(m≠0\)时,\(l_1\)与\(l_2\)的斜率分别为\(m\), \(\dfrac{1}{-m}\),斜率之积 \(m \cdot \dfrac{1}{-m}=-1 \text {, }\),故\(l_1⊥l_2\).
综上,\(l_1⊥l_2\).
(2)由题意知,\(A(0,1)\),\(B(1,0)\), \(A B=\sqrt{2}\),
由\(l_1\)与\(l_2\)相的方程联立方程组 \(\left\{\begin{array}{c} y=m x+1 \\ x=-m y+1 \end{array}\right.\),解得点 \(P\left(\dfrac{1-m}{1+m^{2}}, \dfrac{1+m}{1+m^{2}}\right)\)
则 \(P A=\sqrt{\left(\dfrac{1-m}{1+m^{2}}\right)^{2}+\left(\dfrac{1+m}{1+m^{2}}-1\right)^{2}}=\dfrac{1-m}{\sqrt{1+m^{2}}}\), \(P B=\sqrt{\left(\dfrac{1-m}{1+m^{2}}-1\right)^{2}+\left(\dfrac{1+m}{1+m^{2}}\right)^{2}}=\dfrac{1+m}{\sqrt{1+m^{2}}}\),
由(1)可知\(PA⊥PB\),
\(\therefore S_{\triangle A P B}=\dfrac{1}{2} P A \cdot P B=\dfrac{1-m^{2}}{2\left(1+m^{2}\right)}\),
由四边形\(OAPB\)的面积\(S\)等于两个直角三角形\(OAB\)和\(APB\)的面积之和,
\(\therefore S=\dfrac{1}{2}+\dfrac{1-m^{2}}{2\left(1+m^{2}\right)}=\dfrac{1}{1+m^{2}}\),
故\(m=0\)时,\(S\)有最大值为\(1\).
【B组---提高题】
1.在平面直角坐标平面内有四点\(A(-1,0)\),\(B(2,1)\),\(C(1,5)\),\(D(-2,2)\),\(P\)为该平面内的动点,则\(P\)到\(A\)、\(B\)、\(C\)、\(D\)四点的距离之和的最小值为( )
A. \(10 \sqrt{2}\) \(\qquad \qquad\) B. \(\sqrt{41}+\sqrt{29}\) \(\qquad \qquad\) C. \(14 \sqrt{3}\) \(\qquad \qquad\) D. \(\sqrt{17}+\sqrt{29}\)
2.直线\(l\)通过点\(P(1,3)\)且与两坐标轴的正半轴交于\(A,B\)两点.
(1)直线\(l\)与两坐标轴所围成的三角形面积为\(6\),求直线\(l\)的方程;
(2)求\(OA+OB\)的最小值;
(3)求\(PA⋅PB\)的最小值.
参考答案
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答案 \(D\)
解析 在平面坐标系出描出点\(A,B,C,D\),可得\(P\)为对角线\(AC\)和\(BD\)的交点时,
\(P\)到\(A,B,C,D\)四点的距离之和取得最小值.
由 \(|A C|+|B D|=\sqrt{(1+1)^{2}+(5-0)^{2}}+\sqrt{(2+2)^{2}+(1-2)^{2}}\)\(=\sqrt{29}+\sqrt{17}\).
故选:\(D\).
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答案 (1) \(3x+y-6=0\) (2) \(4+2 \sqrt{3}\) (3) \(6\)
解析 (1)设直线\(l\)的方程为\(y-3=k(x-1)(k<0)\),
由\(x=0\),得\(y=3-k\),由\(y=0\),得 \(x=1-\dfrac{3}{k}\) ,
\(\therefore S_{\triangle A O B}=\dfrac{1}{2}(3-k)\left(1-\dfrac{3}{k}\right)=6\),解得\(k=-3\)
\(∴\)直线\(l\)的方程为\(3x+y-6=0\);
(2) \(O A+O B=3-k+1-\dfrac{3}{k}=4+(-k)+\left(-\dfrac{3}{k}\right)\)\(\geqslant 4+2 \sqrt{-k \cdot\left(-\dfrac{3}{k}\right)}=4+2 \sqrt{3}\).
当且仅当 \(-k=-\dfrac{3}{k}\),即\(k=-\sqrt{3}\)时上式“\(=\)”成立;
(3)设直线\(l\)在\(x,y\)轴上的截距分别为\(a,b\),
则 \((P A \cdot P B)^{2}=\left[(a-1)^{2}+9\right]\left[(b-3)^{2}+1\right]\)
\(=\left[(a-1)^{2}+9\right]\left[\dfrac{9}{(a-1)^{2}}+1\right]=18+(a-1)^{2}+\dfrac{81}{(a-1)^{2}}\)\(\geqslant 18+2 \sqrt{81}=36\).
当且仅当\(a=4\)时上式取等号.
\(∴PA⋅PB\)的最小值为\(6\).
【C组---拓展题】
1.已知\(0<x<2\),\(0<y<2\),则 \(\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+(2-y)^{2}}+\sqrt{(2-x)^{2}+y^{2}}+\sqrt{(2-x)^{2}+(2-y)^{2}}\)
的最小值为( )
A.\(2\sqrt{2}\) \(\qquad \qquad\) B.\(\sqrt{2}\) \(\qquad \qquad\) C.\(2\) \(\qquad \qquad\) D.\(4\sqrt{2}\)
2.已知\(m∈R\),动直线\(l_1:x+my-2=0\)过定点\(A\),动直线\(l_2:mx-y-2m+3=0\)过定点\(B\),若\(l_1\)与\(l_2\)交于点\(P\)(异于点\(A,B\)),则\(|PA|+|PB|\)的最大值为\(\underline{\quad \quad}\).
参考答案
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答案 \(D\)
解析 根据题意,设\(P(x,y)\),\(A(0,0)\),\(B(2,0)\),\(C(2,2)\),\(D(0,2)\),
若\(0<x<2\),\(0<y<2\),则\(P\)在矩形\(ABCD\)的内部,
则 \(\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+(2-y)^{2}}+\sqrt{(2-x)^{2}+y^{2}}+\sqrt{(2-x)^{2}+(2-y)^{2}}\)的几何意义
为\(P\)到\(A、B、C、D\)四点的距离之和,
即 \(\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+(2-y)^{2}}+\sqrt{(2-x)^{2}+y^{2}}+\sqrt{(2-x)^{2}+(2-y)^{2}}=|PA|+|PB|+|PC|+|PD|\),
分析可得,当\(P\)为矩形\(ABCD\)的对角线的交点即\((1,1)\)时,\(|PA|+|PC|\)与\(|PB|+|PD|\)同时取得最小值,
此时\(|PA|+|PB|+|PC|+|PD|\)取得最小值,
且其最小值为 \(|A C|+|B D|=4 \sqrt{2}\);
故选:\(D\).
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答案 \(3\sqrt{2}\)
解析 \(l_1:x+my-2=0\)可变形为\((x-2)+my=0\),令\(y=0\),则\(x=2\),
故动直线\(l_1:x+my-2=0\)过定点\(A(2,0)\),
\(l_2:mx-y-2m+3=0\)可变形为\(m(x-2)-(y-3)=0\),令\(x=2\),则\(y=3\),
故动直线\(l_2:mx-y-2m+3=0\)过定点\(B(2,3)\),
又\(1⋅m+m⋅(-1)=0\),所以直线\(l_1\)与直线\(l_2\)垂直,
则有\(PA⊥PB\),且\(|PA|^2+|PB|^2=|AB|^2=9\),
所以 \(\left(\dfrac{|P A|+|P B|}{2}\right)^{2} \leq \dfrac{|P A|^{2}+|P B|^{2}}{2}=\dfrac{9}{2}\),
即\(|PA|+|PB|≤3\sqrt{2}\),当且仅当\(|PA|=|PB|\)时取等号,
所以\(|PA|+|PB|\)的最大值为\(3\sqrt{2}\).
故答案为:\(3\sqrt{2}\).