Answer:
The given curves are
[tex]y_{1}=(x+c)-1[/tex] and [tex]y=a(x+k)+13[/tex]
Differentiating both the curves with respect to 'x' we get
[tex]\frac{dy_{1}}{dx}=m_{1}=\frac{d(x+c)-1}{dx}\\\\=1\\Similarly\\\frac{dy_{2}}{dx}=m_{2}=\frac{d(ax+ak)+13}{dx}\\\\=a[/tex]
Now for orthogonal trajectories we have product of slopes should be -1
[tex]\Rightarrow m_{1}\times m_{2}=-1\\\\\\1\times a=-1\\\\\therefore a=\frac{-1}{1}=-1[/tex]
