Answer :
To solve this problem, we'll determine the equation of line [tex]\( l_2 \)[/tex] that is perpendicular to line [tex]\( l_1 \)[/tex] and passes through a specific point.
1. Identify the slope of line [tex]\( l_1 \)[/tex]:
The equation of line [tex]\( l_1 \)[/tex] is given as [tex]\( x + 3y = 6 \)[/tex]. First, we'll rewrite this equation in slope-intercept form, [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] represents the slope.
- Start with the equation: [tex]\( x + 3y = 6 \)[/tex].
- Solve for [tex]\( y \)[/tex] by isolating it on one side:
[tex]\[
3y = -x + 6
\][/tex]
[tex]\[
y = -\frac{1}{3}x + 2
\][/tex]
The slope of line [tex]\( l_1 \)[/tex] is [tex]\(-\frac{1}{3}\)[/tex].
2. Find the slope of line [tex]\( l_2 \)[/tex]:
Since line [tex]\( l_2 \)[/tex] is perpendicular to [tex]\( l_1 \)[/tex], its slope will be the negative reciprocal of the slope of line [tex]\( l_1 \)[/tex]. The negative reciprocal of [tex]\(-\frac{1}{3}\)[/tex] is [tex]\(3\)[/tex].
3. Find the point of intersection:
We're given that lines [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] intersect at [tex]\( x = -9 \)[/tex].
To find the corresponding [tex]\( y \)[/tex]-coordinate for this [tex]\( x \)[/tex]-value on line [tex]\( l_1 \)[/tex], substitute [tex]\( x = -9 \)[/tex] into the equation of [tex]\( l_1 \)[/tex]:
[tex]\[
-9 + 3y = 6
\][/tex]
[tex]\[
3y = 15
\][/tex]
[tex]\[
y = 5
\][/tex]
Therefore, the point of intersection is [tex]\((-9, 5)\)[/tex].
4. Write the equation of line [tex]\( l_2 \)[/tex]:
Using the point-slope form of a linear equation, which is [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is a point on the line, we can derive the equation of [tex]\( l_2 \)[/tex]:
- The point is [tex]\((-9, 5)\)[/tex].
- The slope [tex]\( m \)[/tex] is [tex]\( 3 \)[/tex].
Substitute these into the point-slope form:
[tex]\[
y - 5 = 3(x + 9)
\][/tex]
Now, simplify this to get the equation in slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[
y - 5 = 3x + 27
\][/tex]
[tex]\[
y = 3x + 32
\][/tex]
Therefore, the equation of line [tex]\( l_2 \)[/tex] is [tex]\( y = 3x + 32 \)[/tex].
1. Identify the slope of line [tex]\( l_1 \)[/tex]:
The equation of line [tex]\( l_1 \)[/tex] is given as [tex]\( x + 3y = 6 \)[/tex]. First, we'll rewrite this equation in slope-intercept form, [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] represents the slope.
- Start with the equation: [tex]\( x + 3y = 6 \)[/tex].
- Solve for [tex]\( y \)[/tex] by isolating it on one side:
[tex]\[
3y = -x + 6
\][/tex]
[tex]\[
y = -\frac{1}{3}x + 2
\][/tex]
The slope of line [tex]\( l_1 \)[/tex] is [tex]\(-\frac{1}{3}\)[/tex].
2. Find the slope of line [tex]\( l_2 \)[/tex]:
Since line [tex]\( l_2 \)[/tex] is perpendicular to [tex]\( l_1 \)[/tex], its slope will be the negative reciprocal of the slope of line [tex]\( l_1 \)[/tex]. The negative reciprocal of [tex]\(-\frac{1}{3}\)[/tex] is [tex]\(3\)[/tex].
3. Find the point of intersection:
We're given that lines [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] intersect at [tex]\( x = -9 \)[/tex].
To find the corresponding [tex]\( y \)[/tex]-coordinate for this [tex]\( x \)[/tex]-value on line [tex]\( l_1 \)[/tex], substitute [tex]\( x = -9 \)[/tex] into the equation of [tex]\( l_1 \)[/tex]:
[tex]\[
-9 + 3y = 6
\][/tex]
[tex]\[
3y = 15
\][/tex]
[tex]\[
y = 5
\][/tex]
Therefore, the point of intersection is [tex]\((-9, 5)\)[/tex].
4. Write the equation of line [tex]\( l_2 \)[/tex]:
Using the point-slope form of a linear equation, which is [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is a point on the line, we can derive the equation of [tex]\( l_2 \)[/tex]:
- The point is [tex]\((-9, 5)\)[/tex].
- The slope [tex]\( m \)[/tex] is [tex]\( 3 \)[/tex].
Substitute these into the point-slope form:
[tex]\[
y - 5 = 3(x + 9)
\][/tex]
Now, simplify this to get the equation in slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[
y - 5 = 3x + 27
\][/tex]
[tex]\[
y = 3x + 32
\][/tex]
Therefore, the equation of line [tex]\( l_2 \)[/tex] is [tex]\( y = 3x + 32 \)[/tex].
