Câu 11:
Gọi tọa độ chân đường cao kẻ từ A xuống BC là H(x;y)
=>\(AH\perp\)BC
A(-1;2); B(0;3); C(5;-2)
\(\overrightarrow{BC}=\left(5;-5\right);\overrightarrow{BH}=\left(x;y-3\right)\)
\(\overrightarrow{AH}=\left(x+1;y-2\right)\)
B,H,C thẳng hàng nên ta có: \(\dfrac{x}{5}=\dfrac{y-3}{-5}\)
=>x=-y+3
=>x+y=3(1)
AH\(\perp\)BC
=>\(\overrightarrow{AH}\cdot\overrightarrow{BC}=0\)
=>\(5\left(x+1\right)+\left(-5\right)\cdot\left(y-2\right)=0\)
=>\(\left(x+1\right)-\left(y-2\right)=0\)
=>x+1-y+2=0
=>x-y=-3(2)
Từ (1) và (2) ta có hệ phương trình
\(\left\{{}\begin{matrix}x+y=3\\x-y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x+y=3\end{matrix}\right.\)
=>x=0 và y=3
=>Chọn A
Câu 12:
B(-1;3); C(3;1); A(x;y)
\(\overrightarrow{AB}=\left(-1-x;3-y\right)\); \(\overrightarrow{AC}=\left(3-x;1-y\right)\)
\(AB=\sqrt{\left(-1-x\right)^2+\left(3-y\right)^2}=\sqrt{\left(y-3\right)^2+\left(x+1\right)^2}\)
\(AC=\sqrt{\left(3-x\right)^2+\left(1-y\right)^2}=\sqrt{\left(x-3\right)^2+\left(y-1\right)^2}\)
ΔABC vuông cân tại A
=>AB\(\perp\)AC và AB=AC
=>\(\left\{{}\begin{matrix}\overrightarrow{AB}\cdot\overrightarrow{AC}=0\\AB=AC\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(-1-x\right)\left(3-x\right)+\left(3-y\right)\left(1-y\right)=0\\\left(y-3\right)^2+\left(x+1\right)^2=\left(x-3\right)^2+\left(y-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+1\right)\left(x-3\right)+\left(y-3\right)\left(y-1\right)=0\\y^2-6y+9+x^2+2x+1=x^2-6x+9+y^2-2y+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+1\right)\left(x-3\right)+\left(y-3\right)\left(y-1\right)=0\\-6y+2x=-6x-2y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+1\right)\left(x-3\right)+\left(y-3\right)\left(y-1\right)=0\\8x=4y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x\\x^2-2x-3+y^2-4y+3=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x\\x^2-2x+y^2-4y=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x\\x^2-2x+4x^2-4\cdot2x=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x\\5x^2-10x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x\left(x-2\right)=0\\y=2x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(x-2\right)=0\\y=2x\end{matrix}\right.\)
\(x\left(x-2\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Khi x=0 thì \(y=2\cdot0=0\)
Khi x=2 thì \(y=2\cdot2=4\)
=>Chọn B
Câu 13: A(0;4); B(3;4); C(3;0)
\(AB=\sqrt{\left(3-0\right)^2+\left(4-4\right)^2}=3\)
\(AC=\sqrt{\left(3-0\right)^2+\left(0-4\right)^2}=\sqrt{3^2+4^2}=5\)
\(BC=\sqrt{\left(3-3\right)^2+\left(0-4\right)^2}=4\)
Vì \(AB^2+BC^2=AC^2\)
nên ΔABC vuông tại B
=>\(R=\dfrac{AC}{2}=2,5\)
=>Chọn A
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