Assuming that you mean that the equation of line c is
[tex]y=-\dfrac{4}{3}x-4[/tex]
We can deduce that the slope of line c is
[tex]m_c=-\dfrac{4}{3}[/tex]
since the slope is the coefficient multiplying x in the equation of a line. Since we want line d to be perpendicular to line c, their slope must be in the following relationship:
[tex]m_cm_d=-1 \iff m_d=-\dfrac{1}{m_c}=-\dfrac{1}{-\frac{4}{3}}=\dfrac{3}{4}[/tex]
So, line d has slope 3/4 and passes through point (4,-4). Use the equation
[tex]y-y_0=m_d(x-x_0)[/tex]
to find the equation of line d, where
[tex]x_0=4,\quad y_0=-4,\quad m_d=\dfrac{3}{4}[/tex]
