`A(5;-2);B(-1;4)`
`a)` $M\in Oy$
`=>M(0;a)`
`=>\vec{AM}=(-5;a+2)`
`\qquad \vec{BM}=(1;a-4)`
Để $∆MAB$ vuông tại $M$
`=>\vec{AM}.\vec{BM}=0`
`<=>-5.1+(a+2)(a-4)=0`
`<=>-5+a^2-4a+2a-8=0`
`<=>a^2-2a-13=0`
$⇔\left[\begin{array}{l}a=1+\sqrt{14}\\a=1-\sqrt{14}\end{array}\right.$
`=>M(0;1+\sqrt{14})` hoặc `M(0;1-\sqrt{14})`
$\\$
`b)` $K\in Ox$
`=>K(b;0)`
`A(5;-2);B(-1;4)`
`=>\vec{AK}=(b-5;2)`
`\qquad \vec{BK}=(b+1;-4)`
Để $∆KAB$ cân tại $K$
`=>AK=BK`
`=>AK^2=BK^2`
`<=>(b-5)^2+2^2=(b+1)^2+(-4)^2`
`<=>b^2-10b+25+4=b^2+2b+1+16`
`<=>12b=12`
`<=>b=1`
Vậy $K(1;0)$
