The equation of the second line is written in standard form. To know the slope of this line, we can rewrite its equation in slope-intercept form by solving for y.
[tex]\begin{gathered} ax+by=c\Rightarrow\text{ Standard form} \\ y=mx+b\Rightarrow\text{ Slope-intercept form} \\ \text{ Where m is the slope and b is the y-intercept} \end{gathered}[/tex]Then, we have:
[tex]\begin{gathered} 4x-5y=-10 \\ \text{ Subtract 4x from both sides of the equation} \\ 4x-5y-4x=-10-4x \\ -5y=-10-4x \\ \text{Divide by -5 from both sides of the equation} \\ \frac{-5y}{-5}=\frac{-10-4x}{-5} \\ y=\frac{-10}{-5}-\frac{4x}{-5} \\ y=2+\frac{4}{5}x \\ \text{ Reorganize} \\ y=\frac{4}{5}x+2 \\ \text{ Then} \\ $\boldsymbol{m=\frac{4}{5}}$ \end{gathered}[/tex]Now, two lines are perpendicular if their slopes satisfy the following equation:
[tex]\begin{gathered} m_1=-\frac{1}{m_2} \\ \text{ Where }m_1\text{ is the slope of the first equation and} \\ m_2\text{ is the slope of the second equation} \end{gathered}[/tex]In this case, we have:
[tex]\begin{gathered} m_2=\frac{4}{5} \\ m_1=-\frac{1}{\frac{4}{5}_{}} \\ m_1=-\frac{\frac{1}{1}}{\frac{4}{5}_{}} \\ m_1=-\frac{1\cdot5}{1\cdot4} \\ $\boldsymbol{m}_{\boldsymbol{1}}\boldsymbol{=-\frac{5}{4}}$ \end{gathered}[/tex]Step 2Since we already have a point on the line and its slope, then we can use the point-slope formula:
[tex]\begin{gathered} y-y_1=m(x-x_1)\Rightarrow\text{ Point-slope formula} \\ \text{ Where } \\ m\text{ is the slope and} \\ (x_1,y_1)\text{ is a point through which the line passes} \end{gathered}[/tex]Then, we have:
[tex]\begin{gathered} (x_1,y_1)=(6,3) \\ m=-\frac{5}{4} \\ y-3=-\frac{5}{4}(x-6) \\ \text{ Apply the distributive property} \\ y-3=-\frac{5}{4}\cdot x-\frac{5}{4}\cdot-6 \\ y-3=-\frac{5}{4}x+\frac{5}{4}\cdot6 \\ y-3=-\frac{5}{4}x+\frac{30}{4} \\ \text{ Add 3 from both sides of the equation} \\ y-3+3=-\frac{5}{4}x+\frac{30}{4}+3 \\ y=-\frac{5}{4}x+\frac{30}{4}+\frac{12}{4} \\ y=-\frac{5}{4}x+\frac{30+12}{4} \\ y=-\frac{5}{4}x+\frac{42}{4} \\ \text{ Simplify} \\ y=-\frac{5}{4}x+\frac{21\cdot2}{2\cdot2} \\ y=-\frac{5}{4}x+\frac{21}{2} \end{gathered}[/tex]Step 3Therefore, the equation of the line that passes through the point (6,3) that is perpendicular to the line 4x - 5y = -10 is
[tex]$\boldsymbol{y=-\frac{5}{4}x+\frac{21}{2}}$[/tex]