BRAINLIEST 20 POINTS

14.
[tex]y=-\dfrac{5}{8}x+3\ \ \ |\cdot8\\\\8y=-5x+24\ \ \ |+5x\\\\\boxed{5x+8y=24}[/tex]

15.
[tex]k:y=m_1x+b;\ l:y=m_2x+c\\\\k\ ||\ l\ \iff\ m_1=m_2[/tex]

therefore [tex]y=2x+b[/tex]

substitute the coordinates of the point to the equation:

[tex](3;\ 8)\to x=3;\ y=8\\\\8=2\cdot3+b\\8=6+b\ \ \ \ |-6\\b=2[/tex]

Answer: y = 2x + 2

16.
[tex]k:y=m_1x+b;\ l:y=m_2x+c\\\\k\ ||\ l\iff m_1=m_2\\\\k\ \perp\ l\iff m_1\cdot m_2=-1[/tex]

[tex]k:y=-3x+7\\\\l:-2x+6y=3\ \ \ |+2x\\\\6y=2x+3\ \ \ |:6\\\\y=\dfrac{1}{3}x+\dfrac{1}{2}\\\\m_1=-3;\ m_2=\dfrac{1}{3}\\\\m_1\cdot m_2=-3\cdot\dfrac{1}{3}=-1[/tex]

therefore [tex]k\ \perp\ l[/tex]

Answer: perpendicular

17.
a conditions as in the 16th question.

[tex]k:y=-\dfrac{1}{4}x+8\\\\l:-2x+8y=4\ \ \ |+2x\\\\8y=2x+4\ \ \ |:8\\\\y=\dfrac{1}{4}x+\dfrac{1}{2}\\\\m_1=\dfrac{1}{4};\ m_2=\dfrac{1}{4}\\\\m_1=m_2[/tex]

therefore\ [tex]k\ ||\ l[/tex]

Answer: parallel