Để hai đường thẳng vuông góc
\(\Leftrightarrow m\left(4m-5\right)=-1\Leftrightarrow4m^2-5m+1=0\Rightarrow\left[{}\begin{matrix}m=1\\m=\frac{1}{4}\end{matrix}\right.\)
b/ Gọi điểm cố định mà \(d_2\) đi qua là \(M\left(x_0;y_0\right)\)
\(\Rightarrow y_0=\left(4m-5\right)x_0+3m\) \(\forall m\)
\(\Leftrightarrow m\left(4x_0+3\right)-\left(5x_0+y_0\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}4x_0+3=0\\5x_0+y_0=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_0=-\frac{3}{4}\\y_0=\frac{15}{4}\end{matrix}\right.\) \(\Rightarrow M\left(-\frac{3}{4};\frac{15}{4}\right)\)