Question

Cho A(1;-2), B(-3;2). Tìm M thuộc trục tung sao cho 

a, MA + MB nhỏ nhất 

b, \(|MA-MB|\) nhỏ nhất 

Answer 1

Gọi \(M\left(0;m\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AM}=\left(-1;m+2\right)\\\overrightarrow{BM}=\left(3;m-2\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}MA=\sqrt{1+\left(m+2\right)^2}=\sqrt{m^2+4m+5}\\MB=\sqrt{9+\left(m-2\right)^2}=\sqrt{m^2-4m+13}\end{matrix}\right.\)

a.

\(MA+MB=\sqrt{1^2+\left(m+2\right)^2}+\sqrt{3^2+\left(2-m\right)^2}\)

\(MA+MB\ge\sqrt{\left(1+3\right)^2+\left(m+2+2-m\right)^2}=4\sqrt{2}\)

Dấu "=" xảy ra khi \(2-m=3\left(m+2\right)\Leftrightarrow m=-1\)

Hay \(M\left(0;-1\right)\)

b.

\(\left|MA-MB\right|\ge0\)

Dấu "=" xảy ra khi \(MA=MB\Leftrightarrow m^2+4m+5=m^2-4m+13\)

\(\Leftrightarrow m=1\Rightarrow M\left(0;1\right)\)