Respuesta :
We are given the lines
Line 1
[tex]5y=3x+2[/tex]Line 2
[tex]y=\frac{5}{3}x-1[/tex]Line 3
[tex]6x+10y=6[/tex]We want to determine the pair that are parallel, perpendicular or neither
Solution
In order to determine that, we need to obtain the slope of each lines
Note: (1) if the slopes are equal, then they are parallel;
(2) If the product of the slopes equal neagtive one (-1), then they are perpendicular;
(3) Otherwise, they are neither
Now, we will obtain the slope of the lines
We will write each of the equation in this form
[tex]y=mx+c[/tex]Thus, the slope will be the coefficient of x (i.e m)
For Line 1
[tex]\begin{gathered} 5y=3x+2 \\ y=\frac{3}{5}x+\frac{2}{5} \\ \text{Therefore,} \\ m_1=\frac{3}{5} \end{gathered}[/tex]For Line 2
[tex]\begin{gathered} y=\frac{5}{3}x-1 \\ \text{thus,} \\ m_2=\frac{5}{3} \end{gathered}[/tex]For Line 3
[tex]\begin{gathered} 6x+10y=6 \\ 10y=-6x+6 \\ y=-\frac{6}{10}x+\frac{6}{10} \\ y=-\frac{3}{5}x+\frac{3}{5} \\ \text{Therefore,} \\ m_3=-\frac{3}{5} \end{gathered}[/tex]Now, Let us consider Line 1 and Line 2
[tex]\begin{gathered} m_1\ne m_2\text{ (Not parallel)} \\ m_1\times m_2=\frac{3}{5}\times\frac{5}{3}=1\ne-1\text{ (not perpendicular)} \end{gathered}[/tex]Therefore, Line 1 and Line 2 are neither parallel nor perpendicular
For Line 1 and Line 3
[tex]\begin{gathered} m_1\ne m_{3\text{ }}\text{ (not parallel)} \\ m_1\times m_3=\frac{3}{5}\times-\frac{3}{5}=-\frac{9}{25}\ne-1\text{ (not perpendicular)} \end{gathered}[/tex]Therefore, Line 1 and Line 3 are neither parallel nor perpendicular
We consider finally Line 2 and Line 3
[tex]\begin{gathered} m_2\ne m_{3\text{ }}\text{ (not parallel)} \\ m_2\times m_3=\frac{5}{3}\times-\frac{3}{5}=-1\text{ (perpendicular)} \end{gathered}[/tex]Therefore, Line 2 and Line 3 are perpendicular
