Question

Which equation represents a line which is perpendicular to the line 5x+2y=121. y = -5/2x+22. y = 5/2x+33. y = -2/5x-44. y = 2/5x-3

Answer 1

To solve the exercise you can first take the equation of the given line to its slope-intercept form, that is,

`\begin{gathered} y=mx+b \\ \text{ Where m is the slope and} \\ b\text{ is the y-intercept} \end{gathered}`

To take the equation of the given line to its slope-intercept form, you can solve for y, like this

`\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} \\ (2y)/(2)=(12-5x)/(2) \\ y=(12)/(2)-(5x)/(2) \\ y=6-(5)/(2)x \\ \text{ Reordering} \\ y=-(5)/(2)x+6 \end{gathered}`

Now, two lines are perpendicular if their slopes satisfy the equation

`\begin{gathered} m_1=(-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}`

So, in this case, you have

`\begin{gathered} m_1=(-5)/(2) \\ m_2=? \end{gathered}`
`\begin{gathered} m_1=(-1)/(m_2) \\ \text{ Replace and solve for }m_2 \\ (-5)/(2)_{}=(-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} \\ (-5m_2)/(-5)=(-2)/(-5) \\ m_2=(2)/(5) \end{gathered}`

Therefore, the equation that represents a line that is perpendicular to the line 5x + 2y = 12 is

`y=(2)/(5)x-3`