Choose the equation of a line that passes through point (4,-2) and is perpendicular to y=-4x+1.

1 answer

To find the equation of a line that is perpendicular to \( y = -4x + 1 \) and passes through the point \( (4, -2) \), we first need to determine the slope of the given line.

The given line is in the slope-intercept form \( y = mx + b \), where \( m \) is the slope. From the equation \( y = -4x + 1 \), we see that the slope \( m \) is \( -4 \).

For two lines to be perpendicular, the product of their slopes must equal \( -1 \). Therefore, if the slope of the original line is \( -4 \), the slope of the line that is perpendicular to it will be:

\[ m_{\text{perpendicular}} = -\frac{1}{m} = -\frac{1}{-4} = \frac{1}{4} \]

Now, we have the slope of the line we want to find, which is \( \frac{1}{4} \). 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. Plugging in our point \( (4, -2) \) and the slope \( \frac{1}{4} \):

\[ y - (-2) = \frac{1}{4}(x - 4) \]

Simplifying this equation:

\[ y + 2 = \frac{1}{4}(x - 4) \]

\[ y + 2 = \frac{1}{4}x - 1 \]

Now, subtract 2 from both sides to get \( y \) by itself:

\[ y = \frac{1}{4}x - 1 - 2 \]

\[ y = \frac{1}{4}x - 3 \]

Thus, the equation of the line that passes through the point \( (4, -2) \) and is perpendicular to \( y = -4x + 1 \) is:

\[ y = \frac{1}{4}x - 3 \]