Question
Write an equation for line L in point-slope form and slope-intercept form.
L is perpendicular to y equals 3 x.
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y equals 3 xnbsp left parenthesis negative 1 comma negative 3 right parenthesisUpper L
A coordinate system has a horizontal x-axis labeled from negative 6 to 6 in increments of 1 and a vertical y-axis labeled from negative 6 to 6 in increments of 1. From left to right, a line rises in quadrant 3 and passes through the origin into quadrant 1 and is labeled y = 3 x. A second line, labeled L, that falls from left to right is perpendicular to the first line intersecting it at the plotted point (negative 1, negative 3).
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.
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Question content area right
Part 1
Write an equation for line L in point-slope form.
enter your response here
(Simplify your answer. Use integers or fractions for any numbers in the equation.)
Write an equation for line L in slope-intercept form.
enter your response here
(Simplify your answer. Use integers or fractions for any numbers in the equation.)
L is perpendicular to y equals 3 x.
-6
-4
-2
2
4
6
-6
-4
-2
2
4
6
x
y
y equals 3 xnbsp left parenthesis negative 1 comma negative 3 right parenthesisUpper L
A coordinate system has a horizontal x-axis labeled from negative 6 to 6 in increments of 1 and a vertical y-axis labeled from negative 6 to 6 in increments of 1. From left to right, a line rises in quadrant 3 and passes through the origin into quadrant 1 and is labeled y = 3 x. A second line, labeled L, that falls from left to right is perpendicular to the first line intersecting it at the plotted point (negative 1, negative 3).
.
.
.
Question content area right
Part 1
Write an equation for line L in point-slope form.
enter your response here
(Simplify your answer. Use integers or fractions for any numbers in the equation.)
Write an equation for line L in slope-intercept form.
enter your response here
(Simplify your answer. Use integers or fractions for any numbers in the equation.)
Answers
GPT-4o mini
To find the equation for the line \( L \) that is perpendicular to the line given by \( y = 3x \) and passes through the point \((-1, -3)\), we can follow these steps:
### Part 1: Point-Slope Form
1. **Find the slope of the given line**:
The slope of the line \( y = 3x \) is \( m = 3 \).
2. **Find the slope of line \( L \)**:
Since lines \( L \) and \( y = 3x \) are perpendicular, the slope of line \( L \) is the negative reciprocal of \( 3 \). Thus, the slope \( m_L \) of line \( L \) is
\[
m_L = -\frac{1}{3}.
\]
3. **Write the equation in point-slope form**:
The point-slope form of a line is given by
\[
y - y_1 = m(x - x_1)
\]
where \( (x_1, y_1) \) is a point on the line. Here, we use the point \((-1, -3)\), so
\[
y - (-3) = -\frac{1}{3}(x - (-1)).
\]
Simplifying this gives:
\[
y + 3 = -\frac{1}{3}(x + 1).
\]
This is the equation for line \( L \) in **point-slope form**:
\[
y + 3 = -\frac{1}{3}(x + 1).
\]
### Part 2: Slope-Intercept Form
Now we will convert the point-slope form into slope-intercept form (\( y = mx + b \)):
1. **Distribute the slope on the right side**:
\[
y + 3 = -\frac{1}{3}x - \frac{1}{3}.
\]
2. **Isolate \( y \)** by subtracting 3 from both sides:
\[
y = -\frac{1}{3}x - \frac{1}{3} - 3.
\]
3. **Simplify the constant term**:
\[
-3 = -\frac{9}{3} \quad \text{(since } 3 = \frac{9}{3} \text{)}
\]
Therefore,
\[
y = -\frac{1}{3}x - \frac{1}{3} - \frac{9}{3} = -\frac{1}{3}x - \frac{10}{3}.
\]
So the equation for line \( L \) in **slope-intercept form** is:
\[
y = -\frac{1}{3}x - \frac{10}{3}.
\]
### Summary:
- **Point-slope form**: \( y + 3 = -\frac{1}{3}(x + 1) \)
- **Slope-intercept form**: \( y = -\frac{1}{3}x - \frac{10}{3} \)
### Part 1: Point-Slope Form
1. **Find the slope of the given line**:
The slope of the line \( y = 3x \) is \( m = 3 \).
2. **Find the slope of line \( L \)**:
Since lines \( L \) and \( y = 3x \) are perpendicular, the slope of line \( L \) is the negative reciprocal of \( 3 \). Thus, the slope \( m_L \) of line \( L \) is
\[
m_L = -\frac{1}{3}.
\]
3. **Write the equation in point-slope form**:
The point-slope form of a line is given by
\[
y - y_1 = m(x - x_1)
\]
where \( (x_1, y_1) \) is a point on the line. Here, we use the point \((-1, -3)\), so
\[
y - (-3) = -\frac{1}{3}(x - (-1)).
\]
Simplifying this gives:
\[
y + 3 = -\frac{1}{3}(x + 1).
\]
This is the equation for line \( L \) in **point-slope form**:
\[
y + 3 = -\frac{1}{3}(x + 1).
\]
### Part 2: Slope-Intercept Form
Now we will convert the point-slope form into slope-intercept form (\( y = mx + b \)):
1. **Distribute the slope on the right side**:
\[
y + 3 = -\frac{1}{3}x - \frac{1}{3}.
\]
2. **Isolate \( y \)** by subtracting 3 from both sides:
\[
y = -\frac{1}{3}x - \frac{1}{3} - 3.
\]
3. **Simplify the constant term**:
\[
-3 = -\frac{9}{3} \quad \text{(since } 3 = \frac{9}{3} \text{)}
\]
Therefore,
\[
y = -\frac{1}{3}x - \frac{1}{3} - \frac{9}{3} = -\frac{1}{3}x - \frac{10}{3}.
\]
So the equation for line \( L \) in **slope-intercept form** is:
\[
y = -\frac{1}{3}x - \frac{10}{3}.
\]
### Summary:
- **Point-slope form**: \( y + 3 = -\frac{1}{3}(x + 1) \)
- **Slope-intercept form**: \( y = -\frac{1}{3}x - \frac{10}{3} \)