Question

parallel to y=3x and has the point (-2,-1)

Answer 1

To find the equation of a line parallel to \( y = 3x \) that passes through the point \((-2, -1)\), we need to recognize that parallel lines have the same slope.

The given line \( y = 3x \) has a slope of \( 3 \). Therefore, the line we want to find will also have a slope of \( 3 \).

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

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

where \( m \) is the slope and \((x_1, y_1)\) is a point on the line. We have:

- \( m = 3 \)
- \( (x_1, y_1) = (-2, -1) \)

Substituting these values in, we get:

\[
y - (-1) = 3(x - (-2))
\]

This simplifies to:

\[
y + 1 = 3(x + 2)
\]

Now, distributing the \( 3 \):

\[
y + 1 = 3x + 6
\]

Next, we subtract \( 1 \) from both sides to isolate \( y \):

\[
y = 3x + 5
\]

So, the equation of the line that is parallel to \( y = 3x \) and passes through the point \((-2, -1)\) is:

\[
\boxed{y = 3x + 5}
\]