The line \(L_1\) contains points \((-2,1)\) and \((5,-2)\), so the slope \(m_1=\dfrac{-2-1}{5+2}=-\frac{3}{7}\).
So, using the point-slope formula with the point \((-2,1)\) we get \(y-1=-\frac{3}{7}(x+2)\). Solving for \(y\) gives the result \(y=-\frac{3}{7}x+\frac{1}{7}\).
The line \(L_2\) contains the points \((1,1)\) and \((-1,0)\) so the slope \(m_2=\dfrac{0-1}{-1-1}=\frac{1}{2}\). Using the point-slope equation with the point \((-1,0)\) gives the result \(y-0=\frac{1}{2}(x+1)\). Solving for \(y\) gives \(y=\frac{1}{2}x+\frac{1}{2}\).