Question

Which line is perpendicular to the line y=−74x
and goes through the point (−7,5)
?(1 point)
Responses

y=47x+5
y equals Start Fraction 4 over 7 End Fraction x plus 5

y=−74x+5
y equals negative Start Fraction 7 over 4 End Fraction x plus 5

y=47x+9
y equals Start Fraction 4 over 7 End Fraction x plus 9

y=−47x+1

Answer 1

To find a line that is perpendicular to the line \( y = -74x \) and passes through the point \((-7, 5)\), we first need to determine the slope of the given line. The slope of \( y = -74x \) is \(-74\).

The slope of a line that is perpendicular to another line is the negative reciprocal of the slope of the original line. Thus, the slope of the perpendicular line can be calculated as follows:

\[
\text{slope} = -\frac{1}{-74} = \frac{1}{74}
\]

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

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

where \((x_1, y_1)\) is the point \((-7, 5)\) and \(m\) is the slope we found (\(\frac{1}{74}\)). Plugging in these values:

\[
y - 5 = \frac{1}{74}(x - (-7))
\]

Simplifying this:

\[
y - 5 = \frac{1}{74}(x + 7)
\]
\[
y - 5 = \frac{1}{74}x + \frac{7}{74}
\]
\[
y = \frac{1}{74}x + \frac{7}{74} + 5
\]
\[
y = \frac{1}{74}x + \frac{7}{74} + \frac{370}{74}
\]
\[
y = \frac{1}{74}x + \frac{377}{74}
\]

This line in slope-intercept form matches none of the given responses directly, but we can further check which given equations have the slope of \(\frac{1}{74}\).

Now, checking the response options:
1. \(y = 47x + 5\) (slope: 47)
2. \(y = \frac{4}{7}x + 5\) (slope: \(\frac{4}{7}\))
3. \(y = -74x + 5\) (slope: -74)
4. \(y = -\frac{7}{4}x + 5\) (slope: -\(\frac{7}{4}\))
5. \(y = 47x + 9\) (slope: 47)
6. \(y = -47x + 1\) (slope: -47)

None of the provided options appear to directly match our required slope of \(\frac{1}{74}\). It seems that none of those options are correct based on the calculations for a line that is perpendicular to \(y = -74x\) and passes through the point (-7, 5).

Please verify the provided options as they do not reflect the appropriately perpendicular line.