Câu 25 :
\(\overrightarrow{AC}=\left(-5;-5\right)\Rightarrow\overrightarrow{n_p}=\left(5;-5\right)=5\left(1;-1\right)\)
\(\left(AC\right):\left(x-1\right)-\left(y+2\right)=0\)
\(\Rightarrow\left(AC\right):x-y-3=0\)
\(\overrightarrow{BD}=\left(\dfrac{46}{3};-\dfrac{2}{3}\right)\Rightarrow\overrightarrow{n_p}=\left(\dfrac{2}{3};\dfrac{46}{3}\right)=\dfrac{2}{3}\left(1;23\right)\)
\(\left(BD\right):\left(x-15\right)+23\left(y-0\right)=0\)
\(\Rightarrow\left(BD\right):x+23y-15=0\)
\(I\left(x;y\right)=\left(BD\right)\cap\left(AC\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x-y-3=0\\x+23y-15=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow I\left(\dfrac{7}{2};\dfrac{1}{2}\right)\Rightarrow\) Chọn D
Câu 26 :
\(2BD=5DC\Rightarrow\dfrac{BD}{DC}=\dfrac{5}{2}\)
\(\Rightarrow D\) chia \(BC\) theo tỉ lệ \(5:2\), với \(B\) là đầu mút gần hơn
Gọi \(D\left(x_D;y_D\right)\Rightarrow\left\{{}\begin{matrix}x_D=\dfrac{5.x_C+2.x_B}{5+2}=\dfrac{5.3+2.0}{7}=\dfrac{15}{7}\\y_D=\dfrac{5.y_C+2.y_B}{5+2}=\dfrac{5.0+2.1}{7}=\dfrac{2}{7}\end{matrix}\right.\)
\(\Rightarrow I\left(\dfrac{15}{7};\dfrac{2}{7}\right)\Rightarrow\) Chọn A
