Question

select the equations that passes through the point (3, -1) and is parallel to the graph of 5x - 2y = 12. select all that apply
□ $y = \frac{5}{2}x - \frac{17}{2}$
□ $y = -\frac{2}{5}x - \frac{1}{5}$
□ $y = -\frac{5}{2}x + \frac{13}{2}$
□ $y + 3 = \frac{5}{2}(x - 1)$
□ $y + 1 = \frac{5}{2}(x - 3)$
□ $y - 1 = -\frac{2}{5}(x - 3)$

Explanation:

Step1: Find the slope of the given line

First, rewrite the equation \(5x - 2y = 12\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept).
Starting with \(5x-2y = 12\), we solve for \(y\):
\(-2y=-5x + 12\)
\(y=\frac{-5x + 12}{-2}=\frac{5}{2}x-6\)
So the slope \(m\) of the line \(5x - 2y = 12\) is \(\frac{5}{2}\). Since parallel lines have the same slope, the line we are looking for also has a slope of \(\frac{5}{2}\).

Step2: Check each equation

Equation 1: \(y=\frac{5}{2}x-\frac{17}{2}\)

We can check if the point \((3,-1)\) lies on this line. Substitute \(x = 3\) and \(y=-1\) into the equation:
Left - hand side (LHS): \(y=-1\)
Right - hand side (RHS): \(\frac{5}{2}(3)-\frac{17}{2}=\frac{15 - 17}{2}=\frac{-2}{2}=-1\)
Also, the slope is \(\frac{5}{2}\), so this equation is valid.

Equation 2: \(y =-\frac{2}{5}x-\frac{1}{5}\)

The slope of this line is \(-\frac{2}{5}\), which is not equal to \(\frac{5}{2}\), so this line is not parallel. We can also check the point: Substitute \(x = 3\), \(y=-1\)
RHS: \(-\frac{2}{5}(3)-\frac{1}{5}=\frac{-6 - 1}{5}=-\frac{7}{5}
eq - 1\)

Equation 3: \(y=-\frac{5}{2}x+\frac{13}{2}\)

The slope of this line is \(-\frac{5}{2}\), which is not equal to \(\frac{5}{2}\), so this line is not parallel.

Equation 4: \(y + 3=\frac{5}{2}(x - 1)\)

Check if the point \((3,-1)\) lies on this line. Substitute \(x = 3\) and \(y=-1\):
Left - hand side: \(y + 3=-1 + 3=2\)
Right - hand side: \(\frac{5}{2}(3 - 1)=\frac{5}{2}\times2 = 5\)
Since \(2
eq5\), the point \((3,-1)\) does not lie on this line.

Equation 5: \(y + 1=\frac{5}{2}(x - 3)\)

This is the point - slope form \(y - y_1=m(x - x_1)\) where \((x_1,y_1)=(3,-1)\) and \(m = \frac{5}{2}\). So this line has a slope of \(\frac{5}{2}\) and passes through \((3,-1)\).

Equation 6: \(y - 1=-\frac{2}{5}(x - 3)\)

The slope of this line is \(-\frac{2}{5}\), which is not equal to \(\frac{5}{2}\), so this line is not parallel.

Now, let's also check the first equation by substituting \(x = 3\) into \(y=\frac{5}{2}x-\frac{17}{2}\):
\(y=\frac{5}{2}\times3-\frac{17}{2}=\frac{15-17}{2}=\frac{-2}{2}=-1\), which is correct.

Answer:

\(y=\frac{5}{2}x-\frac{17}{2}\), \(y + 1=\frac{5}{2}(x - 3)\)