1.
The parallel lines have the same a slope.
We have:
[tex]y=-\dfrac{3}{2}x+4[/tex]
Therefore, the searched line has the equation:
[tex]y=-\dfrac{3}{2}x+b[/tex]
We know, the line passes through point (4,0). Substitute the coordinates of the point to the equation of the line:
[tex]0=-\dfrac{3}{2}(4)+b\\\\0=-(3)(2)+b\\\\0=-6+b\qquad|+6\\\\6=b\to b=6[/tex]
Answer: [tex]y=-\dfrac{3}{2}x+6[/tex]
2.
Let [tex]k:y=m_1x+b_1[/tex] and [tex]l:y=m_2x+b_2[/tex]
The line l is perpendicular to the line k if and only if [tex]m_1m_2=-1[/tex]
We have [tex]k:y=-\dfrac{1}{3}x-1[/tex]
[tex]l:y=mx+b\\\\l\ \perp\ k\iff-\dfrac{1}{3}m=-1\qquad|\cdot(-3)\\\\m=3[/tex]
The searched line has the equation: [tex]y=3x+b[/tex]
We know, the line passes through point (1,5). Substitute the coordinates of the point to the equation of the line:
[tex]5=3(1)+b\\\\5=3+b\qquad|-3\\\\2=b\to b=2[/tex]
Answer: [tex]y=3x+2[/tex]
