Line

Question

j is shown on the coordinate grid. The table of ordered pairs represents some points on line
k.

x
y
−1
8
0
5
1
2
2
−1
Which system of equations represents line
jand line
k?

Answer 1

To represent the lines with equations, we first need to find the equation for line \( j \) and line \( k \).

Finding the Equation for Line \( k \)

Given the ordered pairs from the table:

• \( (-1, 8) \)
• \( (0, 5) \)
• \( (1, 2) \)
• \( (2, -1) \)

We can find the slope \( m \) using any two points. Let's use \( (0, 5) \) and \( (1, 2) \):

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 5}{1 - 0} = \frac{-3}{1} = -3 \]

Now we can use the point-slope form of the line equation, which is given by:

\[ y - y_1 = m(x - x_1) \]

Using point \( (0, 5) \):

\[ y - 5 = -3(x - 0) \ y - 5 = -3x \ y = -3x + 5 \]

So the equation for line \( k \) is:

\[ y = -3x + 5 \]

Finding the Equation for Line \( j \)

Next, we need more information to determine the equation for line \( j \). If we assume line \( j \) is parallel to line \( k \), it will have the same slope of \( -3 \).

If line \( j \) passes through a specific point, we can use that point to find its equation. However, without the coordinates or more details about line \( j \), we cannot derive its specific equation.

Final System of Equations

If we hypothetically assume that line \( j \) has a slope of \( -3 \) and passes through the point \( (x_1, y_1) \), then its equation could be written as:

\[ y - y_1 = -3(x - x_1) \quad \text{(substituting a specific point for \( (x_1, y_1) \))} \]

Therefore, the general system of equations representing lines \( j \) and \( k \) can be expressed as:

1. \( y = -3x + 5 \) (for line \( k \))
2. \( y = -3x + b \) (for line \( j \) where \( b \) needs to be defined based on the specific point it passes through)

If you have coordinates for line \( j \) or any specific context, please provide those details, and I can give you the exact equation for line \( j \).

Answer 2

(1, 2), (0, -2)

Answer 3

Thank you for providing the points for line \( j \). We'll use the points \( (1, 2) \) and \( (0, -2) \) to find the equation for line \( j \).

Finding the Equation for Line \( j \)

First, we need to calculate the slope \( m \) of line \( j \) using the points \( (1, 2) \) and \( (0, -2) \):

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 2}{0 - 1} = \frac{-4}{-1} = 4 \]

Now that we have the slope, we can use the point-slope form of the line equation. Let's use point \( (0, -2) \):

\[ y - y_1 = m(x - x_1) \]

Substituting the point and the slope:

\[ y - (-2) = 4(x - 0) \ y + 2 = 4x \ y = 4x - 2 \]

So the equation for line \( j \) is:

\[ y = 4x - 2 \]

Line \( k \)

As previously determined, the equation for line \( k \) is:

\[ y = -3x + 5 \]

Final System of Equations

Now, the system of equations representing lines \( j \) and \( k \) is:

1. \( y = 4x - 2 \) (for line \( j \))
2. \( y = -3x + 5 \) (for line \( k \))

Thus, the system of equations is:

\[ \begin{cases} y = 4x - 2 \ y = -3x + 5 \end{cases} \]