Answer
a) $y=-\dfrac{2}{3}x-\dfrac{7}{3}$
b) $y=\dfrac{3}{2}x+15$
Work Step by Step
We are given:
Line $d$: $2x+3y=5$
$(-8,3)$
Determine the slope $m_d$ of the given line:
$2x-5=-3y$
$y=-\dfrac{2}{3}x+\dfrac{5}{3}$
$m_d=-\dfrac{2}{3}$
a) Let $d_1$ be a line parallel to line $d$. We have:
$m_1=m_d=-\dfrac{2}{3}$
Determine the line $d_1$ using the slope $m_1$ and the point $(-8,3)$:
$y-3=-\dfrac{2}{3}(x+8)$
$y-3=-\dfrac{2}{3}x-\dfrac{16}{3}$
$y=-\dfrac{2}{3}x-\dfrac{16}{3}+3$
$y=-\dfrac{2}{3}x-\dfrac{7}{3}$
b) Let $d_2$ be a line perpendicular to the line $d$. We have:
$m_2\cdot m_d=-1$
$m_2\cdot \left(-\dfrac{2}{3}\right)=-1$
$m_2=\dfrac{3}{2}$
Determine the line $d_2$ using the slope $m_2$ and the point $(-8,3)$:
$y-3=\dfrac{3}{2}(x+8)$
$y-3=\dfrac{3}{2}x+12$
$y=\dfrac{3}{2}x+12+3$
$y=\dfrac{3}{2}x+15$
