Question

in each case, determine k so that the line is parallel to the line (2x - 4y + 26 = 0)
(a) (4x + ky = 0)
(b) (kx - 2y - 4 = 0)
(a) what value of k will make (4x + ky = 0) parallel to (2x - 4y + 26 = 0)?
k = (\boxed{quad}) (simplify your answer. type an integer or a fraction.)
(b) what value of k will make (kx - 2y - 4 = 0) parallel to (2x - 4y + 26 = 0)?
k = (\boxed{quad}) (simplify your answer. type an integer or a fraction.)

Explanation:

Step1: Recall parallel line condition

For lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, parallelism requires $\frac{a_1}{a_2} = \frac{b_1}{b_2}
eq \frac{c_1}{c_2}$.

Step2: Solve for k in part (a)

Given lines $4x + ky - 8 = 0$ and $2x - 4y + 28 = 0$. Set $\frac{4}{2} = \frac{k}{-4}$
$\frac{4}{2} = \frac{k}{-4} \implies 2 = \frac{k}{-4} \implies k = 2 \times (-4) = -8$
Verify $\frac{-8}{28}
eq 2$, which holds.

Step3: Solve for k in part (b)

Given lines $kx - 2y - 4 = 0$ and $2x - 4y + 28 = 0$. Set $\frac{k}{2} = \frac{-2}{-4}$
$\frac{k}{2} = \frac{-2}{-4} \implies \frac{k}{2} = \frac{1}{2} \implies k = 1$
Verify $\frac{-4}{28}
eq \frac{1}{2}$, which holds.

Answer:

(a) $k = -8$
(b) $k = 1$