Question

What value completes the equation of the line that is perpendicular to the line y=−34x+7
and goes through the point (6,1)
?(1 point)
The line that is perpendicular to y=−34x+7
and passes through the point (6,1)
is y=43x+
.
Remaining Attempts : 3

Answer 1

To find the equation of a line that is perpendicular to the line \( y = -34x + 7 \) and passes through the point \( (6, 1) \), we need to determine the slope of the perpendicular line.

1. **Identify the slope of the given line:**
The slope (\( m \)) of the given line \( y = -34x + 7 \) is \( -34 \).

2. **Determine the slope of the perpendicular line:**
The slope of a line that is perpendicular to another line is the negative reciprocal of the slope of the original line. Therefore, the slope of the perpendicular line is:
\[
m_{\text{perpendicular}} = -\frac{1}{-34} = \frac{1}{34}
\]

3. **Use the point-slope form of a line:**
We can use the point-slope form of the equation of a line, which is given by:
\[
y - y_1 = m(x - x_1)
\]
where \( (x_1, y_1) \) is the point the line passes through and \( m \) is the slope. For our case, \( (x_1, y_1) = (6, 1) \) and \( m = \frac{1}{34} \).
Substituting these values into the equation, we have:
\[
y - 1 = \frac{1}{34}(x - 6)
\]

4. **Simplify to the slope-intercept form (y = mx + b):**
Distributing the slope:
\[
y - 1 = \frac{1}{34}x - \frac{6}{34}
\]
Simplifying \( -\frac{6}{34} \) gives \( -\frac{3}{17} \):
\[
y - 1 = \frac{1}{34}x - \frac{3}{17}
\]
Now, add 1 (which is \( \frac{17}{17} \)) to both sides to solve for \( y \):
\[
y = \frac{1}{34}x - \frac{3}{17} + 1
\]
\[
y = \frac{1}{34}x - \frac{3}{17} + \frac{17}{17}
\]
\[
y = \frac{1}{34}x + \frac{14}{17}
\]

5. **Final result:**
The equation of the line is
\[
y = \frac{1}{34}x + \frac{14}{17}
\]

Now, the value that completes the equation in the form \( y = \frac{1}{34}x + b \) is \( b = \frac{14}{17} \).

So, the final answer is:

The line that is perpendicular to \( y = -34x + 7 \) and passes through the point \( (6, 1) \) is \( y = \frac{1}{34}x + \frac{14}{17} \).