As the first line passes through the origin, that is point (0,0) and the through the point (4,-5), you can find the slope of this line with the formula
[tex]\begin{gathered} m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \text{ Where m is the slope of the line and} \\ (x_1,y_1),(x_2,y_2)\text{ are two points through which the line passes} \end{gathered}[/tex]
So, you have
[tex]\begin{gathered} (x_1,y_1)=(0,0) \\ (x_2,y_2)=(4,-5) \\ m=\frac{-5-0}{4-0} \\ m=\frac{-5}{4} \end{gathered}[/tex]
Now, you can use the point-slope formulas to find the equation of the line in its slope-intercept form
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y-0_{}=\frac{-5}{4}(x-0) \\ y=\frac{-5}{4}x \end{gathered}[/tex]
Therefore, the equation of the first line and the answer of numeral b) is
[tex]y=\frac{-5}{4}x[/tex]
On the other hand, the slopes of perpendicular lines satisfy the following equation
[tex]\begin{gathered} m_2=\frac{-1}{m_1} \\ \text{ Where }m_1\text{ is the slope of line 1 and} \\ m_2\text{ is the slope of line 2} \end{gathered}[/tex]
So, to find the equation of the second line that is perpendicular to the first, you can find its slope and then use the point-slope formula
[tex]\begin{gathered} m_1=\frac{-5}{4}_{} \\ m_2=\frac{-1}{\frac{-5}{4}_{}} \\ m_2=\frac{\frac{-1}{1}}{\frac{-5}{4}_{}} \\ m_2=\frac{-1\cdot4}{1\cdot-5} \\ m_2=\frac{4}{5} \end{gathered}[/tex]
Now, using the point-slope formula with the point (4,-5)
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y-(-5)_{}=\frac{4}{5}(x-4) \\ y+5_{}=\frac{4}{5}x-\frac{16}{5} \\ \text{ Subtract 5 from both sides of the equation} \\ y+5-5_{}=\frac{4}{5}x-\frac{16}{5}-5 \\ y=\frac{4}{5}x-\frac{41}{5} \end{gathered}[/tex]
Therefore, the equation of the second line and the answer of numeral a) is
[tex]y=\frac{4}{5}x-\frac{41}{5}[/tex]
Graphically,
