Câu 15: M thuộc trục Oy nên M(0;y)
A(-2;-3); B(4;7); M(0;y)
\(\overrightarrow{AB}=\left(4+2;7+3\right)=\left(6;10\right)\)
\(\overrightarrow{AM}=\left(0+2;y+3\right)=\left(2;y+3\right)\)
Để A,B,M thẳng hàng thì \(\dfrac{y+3}{10}=\dfrac{2}{6}=\dfrac{1}{3}\)
=>\(y+3=\dfrac{10}{3}\)
=>\(y=\dfrac{1}{3}\)
=>\(M\left(0;\dfrac{1}{3}\right)\)
Câu 17: A(2;-4); B(6;0); C(m;4)
\(\overrightarrow{AB}=\left(6-2;0+4\right)\)
=>\(\overrightarrow{AB}=\left(4;4\right)\)
\(\overrightarrow{AC}=\left(m-2;4+4\right)\)
=>\(\overrightarrow{AC}=\left(m-2;8\right)\)
Để A,B,C thẳng hàng thì \(\dfrac{m-2}{4}=\dfrac{8}{4}\)
=>m-2=8
=>m=10
=>Chọn A
Câu 18: M thuộc trục x'Ox nên M(x;0)
A(0;-2); B(-3;1); M(x;0)
\(\overrightarrow{AB}=\left(-3-0;1+2\right)=\left(-3;3\right)=\left(-1;1\right)\)
\(\overrightarrow{AM}=\left(x-0;0+2\right)=\left(x;2\right)\)
A,B,M thẳng hàng
=>\(\dfrac{x}{-1}=\dfrac{2}{1}=2\)
=>x=-2
=>M(-2;0)
=>Chọn A
Câu 19: A(1;-1); B(2;4); C(-2;-7); D(3;3)
\(\overrightarrow{AB}=\left(2-1;4+1\right)=\left(1;5\right)\)
\(\overrightarrow{AC}=\left(-2-1;-7+1\right)=\left(-3;-6\right)\)
=>\(\overrightarrow{CA}=\left(3;6\right)=\left(1;2\right)\)
\(\overrightarrow{AD}=\left(3-1;3+1\right)=\left(2;6\right)\)
\(\overrightarrow{BC}=\left(-2-2;-7-4\right)=\left(-4;-11\right)\)
\(\overrightarrow{BD}=\left(3-2;3-4\right)=\left(1;1\right)\)
\(\overrightarrow{CD}=\left(3+2;3+7\right)=\left(5;10\right)\)
Vì \(\dfrac{1}{5}=\dfrac{2}{10}\)
nên C,A,D thẳng hàng
=>Chọn D
