Question

On a coordinate plane, line D F goes through points (negative 1, negative 3) and (2, 3). Point G is at (negative 4, negative 4).
Which point on the y-axis lies on the line that passes through point G and is parallel to line DF?

(–2, 0)
(0, –2)
(0, 4)
(4, 0)

Answer 1

To find the point on the y-axis that lies on the line passing through point G and is parallel to line DF, we first need to determine the slope of line DF.

The slope \( m \) of a line through points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:

\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

For line DF, we have the points \((-1, -3)\) and \((2, 3)\).

Calculating the slope \( m \):

\[
m = \frac{3 - (-3)}{2 - (-1)} = \frac{3 + 3}{2 + 1} = \frac{6}{3} = 2
\]

The slope of line DF is 2. Since we need to find a line that is parallel to line DF, it will also have a slope of 2.

Next, we will use point G at \((-4, -4)\) to find the equation of the line that has a slope of 2 and passes through point G. Using the point-slope form of the equation of a line:

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

where \((x_1, y_1)\) is point G \((-4, -4)\) and \(m = 2\):

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

Simplifying this equation:

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

\[
y + 4 = 2x + 8
\]

\[
y = 2x + 8 - 4
\]

\[
y = 2x + 4
\]

Now, we need to find where this line intersects the y-axis. The y-axis corresponds to \(x = 0\).

Substituting \(x = 0\) into the equation \(y = 2x + 4\):

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

Thus, the point on the y-axis is \((0, 4)\).

Therefore, the point on the y-axis that lies on the line that passes through point G and is parallel to line DF is:

\[
\boxed{(0, 4)}
\]