Respuesta :
To solve the exercise you can first take the equation of the given line to its slope-intercept form, that is,
[tex]\begin{gathered} y=mx+b \\ \text{ Where m is the slope and} \\ b\text{ is the y-intercept} \end{gathered}[/tex]To take the equation of the given line to its slope-intercept form, you can solve for y, like this
[tex]\begin{gathered} 5x+2y=12 \\ \text{ Subtract 5x from both sides of the equation} \\ 5x+2y-5x=12-5x \\ 2y=12-5x \\ \text{ Divide by 2 into both sides of the equation} \\ \frac{2y}{2}=\frac{12-5x}{2} \\ y=\frac{12}{2}-\frac{5x}{2} \\ y=6-\frac{5}{2}x \\ \text{ Reordering} \\ y=-\frac{5}{2}x+6 \end{gathered}[/tex]
Now, two lines are perpendicular if their slopes satisfy the equation
[tex]\begin{gathered} m_1=\frac{-1}{m_2} \\ \text{ Where }m_1\text{ is the slope of the first line and} \\ m_2\text{ is the slope of the second line} \end{gathered}[/tex]So, in this case, you have
[tex]\begin{gathered} m_1=\frac{-5}{2} \\ m_2=? \end{gathered}[/tex][tex]\begin{gathered} m_1=\frac{-1}{m_2} \\ \text{ Replace and solve for }m_2 \\ \frac{-5}{2}_{}=\frac{-1}{m_2} \\ \text{ Apply cross multiplication} \\ -5\cdot m_2=-1\cdot2 \\ -5m_2=-2 \\ \text{ Divide by -5 into both sides of the equation} \\ \frac{-5m_2}{-5}=\frac{-2}{-5} \\ m_2=\frac{2}{5} \end{gathered}[/tex]Therefore, the equation that represents a line that is perpendicular to the line 5x + 2y = 12 is
[tex]y=\frac{2}{5}x-3[/tex]