We have the following:
A.
First we find the slope of the line with the following points:
(0, 3) and (5,0)
[tex]m=\frac{0-3}{5-0}=-\frac{3}{5}[/tex]now, for b, with the point (0,3)
[tex]\begin{gathered} 3=-\frac{3}{5}\cdot0+b \\ b=3 \end{gathered}[/tex]The equation is:
[tex]y=-\frac{3}{5}x+3[/tex]B.
The area is:
[tex]\begin{gathered} A=\frac{AC\cdot CB}{2} \\ A=\frac{3\cdot5}{2}=\frac{15}{2} \\ A=7.5 \end{gathered}[/tex]The area is 7.5 square units
for, perimeter:
[tex]\begin{gathered} p=AC+CB+AB \\ AB^2=AC^2+CB^2 \\ AB^2=3^2+5^2=9+25=34 \\ AB=\sqrt[]{34} \\ p=3+5+\sqrt[]{34} \\ p=13.83 \end{gathered}[/tex]The perimeter is 13.83 units
C.
when two lines are perpendicular they fulfill the following
[tex]m_1\cdot m_2=-1[/tex]therefore,
[tex]\begin{gathered} -\frac{3}{5}\cdot m_2=-1 \\ m_2=\frac{5}{3} \end{gathered}[/tex]Therefore, the equation is:
[tex]y=\frac{5}{3}x+3[/tex]