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Question
in each case, determine k so that the line is parallel to the line $4x - 2y + 25 = 0$
(a) $5x + ky = 20$
(b) $kx - 8y - 32 = 0$
(a) what value of k will make $5x + ky = 20$ parallel to $4x - 2y + 25 = 0$
$k = \square$ (simplify your answer. type an integer or a fraction.)
(b) what value of k will make $kx - 8y - 32 = 0$ parallel to $4x - 2y + 25 = 0$
$k = \square$ (simplify your answer. type an integer or a fraction.)
Step1: Find slope of given line
Rewrite $4x - 2y + 25 = 0$ to slope-intercept form $y=mx+b$:
$-2y = -4x -25$
$y = 2x + \frac{25}{2}$
Slope $m_1 = 2$
Parallel lines have equal slopes.
Step2: Solve for k in part (a)
Rewrite $5x + ky = 20$:
$ky = -5x + 20$
$y = -\frac{5}{k}x + \frac{20}{k}$
Set slope equal to 2:
$-\frac{5}{k} = 2$
Solve for k:
$k = -\frac{5}{2}$
Step3: Solve for k in part (b)
Rewrite $kx - 8y - 32 = 0$:
$-8y = -kx + 32$
$y = \frac{k}{8}x - 4$
Set slope equal to 2:
$\frac{k}{8} = 2$
Solve for k:
$k = 16$
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(a) $k = -\frac{5}{2}$
(b) $k = 16$