Write an equation of the line that passes through (−3,−1)

and is perpendicular to the line y=−1/9x−6
.(1 point)

1 answer

To find the equation of a line that is perpendicular to a given line, we need to determine the slope of the given line and then use the negative reciprocal of that slope for our new line.

The given line is:

\[ y = -\frac{1}{9}x - 6 \]

From this equation, we can see that the slope (m) of the line is:

\[ m = -\frac{1}{9} \]

To find the slope of the line that is perpendicular to it, we take the negative reciprocal of \( -\frac{1}{9} \):

\[ m_{\text{perpendicular}} = -\frac{1}{-1/9} = 9 \]

Now we have the slope of the perpendicular line, which is \( 9 \). We also have a point that the line passes through: \( (-3, -1) \).

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

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

where \((x_1, y_1)\) is the point the line passes through and \(m\) is the slope. Substituting our values:

  • \(x_1 = -3\)
  • \(y_1 = -1\)
  • \(m = 9\)

The equation becomes:

\[ y - (-1) = 9(x - (-3)) \]

This simplifies to:

\[ y + 1 = 9(x + 3) \]

Distributing \(9\):

\[ y + 1 = 9x + 27 \]

Next, we isolate \(y\) by subtracting \(1\) from both sides:

\[ y = 9x + 27 - 1 \]

Finally, we simplify:

\[ y = 9x + 26 \]

Thus, the equation of the line that passes through \((-3, -1)\) and is perpendicular to the line \(y = -\frac{1}{9}x - 6\) is:

\[ y = 9x + 26 \]