Given that a = (3, -5) and b = (-5, -15), find 1/4 (b + a)

For A(1,2):

$$A' = (-2, 1)$$

For B(4,2):

$$B' = (-2, 4)$$

For C(1,4):

$$C' = (-4, 1)$$

The coordinates of the image are $A'(-2,1)$, $B'(-2,4)$, and $C'(-4,1)$.
(Since I cannot draw on the provided grid, I am providing the coordinates of the image.)
The final answer is $\boxed{A'(-2,1), B'(-2,4), C'(-4,1)}$.

8. Given that $\underline{a} = \begin{pmatrix} 3 \\ -5 \end{pmatrix}$ and $\underline{b} = \begin{pmatrix} 5 \\ -15 \end{pmatrix}$, find $\frac{1}{4}(\underline{b} + \underline{a})$.

$$\underline{b} + \underline{a} = \begin{pmatrix} 5 \\ -15 \end{pmatrix} + \begin{pmatrix} 3 \\ -5 \end{pmatrix}$$ $$\underline{b} + \underline{a} = \begin{pmatrix} 5+3 \\ -15+(-5) \end{pmatrix}$$ $$\underline{b} + \underline{a} = \begin{pmatrix} 8 \\ -20 \end{pmatrix}$$

$$\frac{1}{4}(\underline{b} + \underline{a}) = \frac{1}{4} \begin{pmatrix} 8 \\ -20 \end{pmatrix}$$ $$\frac{1}{4}(\underline{b} + \underline{a}) = \begin{pmatrix} \frac{1}{4} \times 8 \\ \frac{1}{4} \times (-20) \end{pmatrix}$$ $$\frac{1}{4}(\underline{b} + \underline{a}) = \begin{pmatrix} 2 \\ -5 \end{pmatrix}$$

The final answer is $\boxed{\begin{pmatrix} 2 \\ -5 \end{pmatrix}}$.