First, the slopes of two perpendicular lines satisfy
[tex]\begin{gathered} m_1=\frac{-1}{m_2} \\ \text{ Where } \\ m_1=\text{ slope of line 1} \\ m_2=\text{ slope of line }2 \end{gathered}[/tex]Then, if you take Y=-3/4x as line 1, you have
[tex]\begin{gathered} m_1=\frac{-3}{4} \\ \text{ Using the formula above} \\ \frac{-3}{4}=\frac{-1}{m_2} \\ \text{Multiply both sides of equation by m2} \\ m_2\cdot\frac{-3}{4}=\frac{-1}{m_2}\cdot m_2 \\ _{}m_2\cdot\frac{-3}{4}=-1 \\ \text{ Multiply both sides of equation by }-\frac{4}{3} \\ m_2\cdot\frac{-3}{4}\cdot\frac{-4}{3}=-1\cdot\frac{-4}{3} \\ m_2\cdot\frac{12}{12}=\frac{4}{3} \\ m_2=\frac{4}{3} \end{gathered}[/tex]Now, using the point slope formula you have
[tex]\begin{gathered} y-y_2=m_2(x-x_2) \\ y-3=\frac{4}{3}(x-3) \\ y-3=\frac{4}{3}x-\frac{12}{3} \\ y-3=\frac{4}{3}x-4 \\ \text{ Add 3 to both sides of the eqution} \\ y-3+3=\frac{4}{3}x-4+3 \\ y=\frac{4}{3}x-1 \end{gathered}[/tex]Therefore, the slope-intercept form of the equation of the line described is
[tex]y=\frac{4}{3}x-1[/tex]
