Question
Write an equation for line L in point-slope form and slope-intercept form.
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.)
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.)
Answers
GPT-4o mini
To find the equation of line \( L \) in point-slope form, we need to find the slope of line \( L \) and use the point it passes through, which is \( (-1, -3) \).
First, we know that line \( y = 3x \) has a slope of \( 3 \). Since line \( L \) is perpendicular to line \( y = 3x \), its slope will be the negative reciprocal of \( 3 \).
The negative reciprocal of \( 3 \) is:
\[
m = -\frac{1}{3}
\]
Now 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. Substituting the values \( (x_1, y_1) = (-1, -3) \) and \( m = -\frac{1}{3} \):
\[
y - (-3) = -\frac{1}{3}(x - (-1))
\]
This simplifies to:
\[
y + 3 = -\frac{1}{3}(x + 1)
\]
This is the equation of line \( L \) in point-slope form.
### So the answer is:
\[
y + 3 = -\frac{1}{3}(x + 1)
\]
First, we know that line \( y = 3x \) has a slope of \( 3 \). Since line \( L \) is perpendicular to line \( y = 3x \), its slope will be the negative reciprocal of \( 3 \).
The negative reciprocal of \( 3 \) is:
\[
m = -\frac{1}{3}
\]
Now 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. Substituting the values \( (x_1, y_1) = (-1, -3) \) and \( m = -\frac{1}{3} \):
\[
y - (-3) = -\frac{1}{3}(x - (-1))
\]
This simplifies to:
\[
y + 3 = -\frac{1}{3}(x + 1)
\]
This is the equation of line \( L \) in point-slope form.
### So the answer is:
\[
y + 3 = -\frac{1}{3}(x + 1)
\]