(d): y=(m-2)x+3
=>(m-2)x-y+3=0
=>Vecto pháp tuyến là \(\overrightarrow{a}=\left(m-2;-1\right)\)
=>\(\left|\overrightarrow{a}\right|=\sqrt{\left(m-2\right)^2+\left(-1\right)^2}=\sqrt{\left(m-2\right)^2+1}\)
(d'): y=x-4
=>x-y-4=0
=>Vecto pháp tuyến là \(\overrightarrow{b}=\left(1;-1\right)\)
=>\(\left|\overrightarrow{b}\right|=\sqrt{1^2+\left(-1\right)^2}=\sqrt2\)
\(\hat{\left(d\right);\left(d^{\prime}\right)}=45^0\)
=>\(\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{\left|\overrightarrow{a}\right|\cdot\left|\overrightarrow{b}\right|}=cos45=\frac{\sqrt2}{2}\)
=>\(\frac{1\left(m-2\right)+\left(-1\right)\cdot\left(-1\right)}{\sqrt{\left(m-2\right)^2+1}\cdot\sqrt2}=\frac{\sqrt2}{2}\)
=>\(\frac{m-2+1}{\sqrt{\left(m-2\right)^2+1}}=1\)
=>\(\sqrt{\left(m-2\right)^2+1}=m-1\)
=>\(\begin{cases}m-1\ge0\\ \left(m-2\right)^2+1=\left(m-1\right)^2\end{cases}\Rightarrow\begin{cases}m\ge1\\ m^2-4m+4+1=m^2-2m+1\end{cases}\)
=>\(\begin{cases}m\ge1\\ -4m+5=-2m+1\end{cases}\Rightarrow\begin{cases}m\ge1\\ -2m=-4\end{cases}\Rightarrow m=2\)
