d A line with a gradient of $-\frac{2}{5}$ and passing through $D(15,-9)$.
\(\boxed{y = -\frac{2}{5}x - 3}\)
Step 1 :Given the gradient $m = -\frac{2}{5}$ and the point $D(15, -9)$
Step 2 :Use the point-slope form of a linear equation: $y - y_1 = m(x - x_1)$
Step 3 :Plug in the values: $y - (-9) = -\frac{2}{5}(x - 15)$
Step 4 :Simplify the equation: $y = -\frac{2}{5}x - 3$
Step 5 :\(\boxed{y = -\frac{2}{5}x - 3}\)