Question

Find the equation of the line that is parallel to the line y=25x+1 and passes through the point (5,5) .(1 point) The line that is parallel to y=25x+1 and passes through the point (5,5) is y=25x+blank.

Answer 1

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+blank.

Answer 2

To find the equation of the line that is parallel to \( y = 25x + 1 \) and passes through the point \( (5, 5) \), we start by recognizing that parallel lines have the same slope. The slope of the given line \( y = 25x + 1 \) is \( 25 \).

Next, we can 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 a point on the line and \( m \) is the slope. Here, \( (x_1, y_1) = (5, 5) \) and \( m = 25 \).

Plugging in the values:

\[
y - 5 = 25(x - 5)
\]

Now, simplify the equation:

\[
y - 5 = 25x - 125
\]

Add \( 5 \) to both sides:

\[
y = 25x - 120
\]

Thus, the line that is parallel to \( y = 25x + 1 \) and passes through the point \( (5, 5) \) is:

\[
y = 25x - 120
\]

So, the blank in your equation \( y = 25x + \text{blank} \) is \( -120 \).

Final answer: The line is \( y = 25x - 120 \).

Answer 3

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

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

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

3. **Use the point-slope form of the line equation**:
The point-slope form is given by:
\[
y - y_1 = m(x - x_1)
\]
where \( (x_1, y_1) = (6, 1) \) and \( m = \frac{1}{34} \).

Substituting in the values, we have:
\[
y - 1 = \frac{1}{34}(x - 6)
\]

4. **Simplify the equation**:
Distributing the slope on the right side:
\[
y - 1 = \frac{1}{34}x - \frac{6}{34}
\]
\[
y - 1 = \frac{1}{34}x - \frac{3}{17}
\]

Now, add \( 1 \) (which is \( \frac{17}{17} \)) to both sides:
\[
y = \frac{1}{34}x - \frac{3}{17} + \frac{17}{17}
\]

Combine the constant terms on the right:
\[
y = \frac{1}{34}x + \left(\frac{17}{17} - \frac{3}{17}\right)
\]
\[
y = \frac{1}{34}x + \frac{14}{17}
\]

Now, we want to write the equation in the desired form \( y = 43x + \text{blank} \). Notice that \( \frac{1}{34} = \frac{43}{34} \), so we can express this as:
\[
y = \frac{43}{34}x + \frac{14}{17}
\]

However, this won't match your requested format exactly. Instead, let's convert \( \frac{14}{17} \) into a decimal or a more manageable fraction.

5. **Finding the blank**:
Since \( y = 43x + \frac{14}{17} \), the blank in your original format \( y = 43x + \text{blank} \) corresponds to \( \frac{14}{17} \).

To summarize, the line that is perpendicular to \( y = -34x + 7 \) and passes through \( (6, 1) \) can be expressed as:
\[
y = 43x + \frac{14}{17}
\]

So, the value that completes the equation is \( \frac{14}{17} \).