Answer
a) $y=\dfrac{5}{4}x-\dfrac{23}{4}$
b) $y=-\dfrac{4}{5}x+\dfrac{2}{5}$
Work Step by Step
We are given:
Line $d$: $5x-4y=8$
$(3,-2)$
Determine the slope $m_d$ of the given line:
$5x-8=4y$
$y=\dfrac{5}{4}x-2$
$m_d=\dfrac{5}{4}$
a) Let $d_1$ be a line parallel to line $d$. We have:
$m_1=m_d=\dfrac{5}{4}$
Determine the line $d_1$ using the slope $m_1$ and the point $(3,-2)$:
$y-(-2)=\dfrac{5}{4}(x-3)$
$y+2=\dfrac{5}{4}x-\dfrac{15}{4}$
$y=\dfrac{5}{4}x-\dfrac{15}{4}-2$
$y=\dfrac{5}{4}x-\dfrac{23}{4}$
b) Let $d_2$ be a line perpendicular to the line $d$. We have:
$m_2\cdot m_d=-1$
$m_2\cdot \dfrac{5}{4}=-1$
$m_2=-\dfrac{4}{5}$
Determine the line $d_2$ using the slope $m_2$ and the point $(3,-2)$:
$y-(-2)=-\dfrac{4}{5}(x-3)$
$y+2=-\dfrac{4}{5}x+\dfrac{12}{5}$
$y=-\dfrac{4}{5}x+\dfrac{12}{5}-2$
$y=-\dfrac{4}{5}x+\dfrac{2}{5}$