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To determine which of the given equations is perpendicular to the graph of the equation [tex]\(-2x + 3y = 6\)[/tex], we need to follow these steps:
1. Find the slope of the given equation:
We start by rewriting [tex]\(-2x + 3y = 6\)[/tex] in the slope-intercept form of [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] represents the slope.
[tex]\[ -2x + 3y = 6 \][/tex]
Add [tex]\(2x\)[/tex] to both sides:
[tex]\[ 3y = 2x + 6 \][/tex]
Divide both sides by 3:
[tex]\[ y = \frac{2}{3}x + 2 \][/tex]
So, the slope [tex]\(m\)[/tex] of the given line is [tex]\(\frac{2}{3}\)[/tex].
2. Determine the slope of the perpendicular line:
Lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex]. Therefore, if one line has a slope [tex]\(m\)[/tex], the perpendicular line will have a slope of [tex]\(-\frac{1}{m}\)[/tex].
For our slope, [tex]\(\frac{2}{3}\)[/tex], the perpendicular slope would be:
[tex]\[ -\frac{1}{\left(\frac{2}{3}\right)} = -\frac{3}{2} \][/tex]
3. Compare the slopes of the given options:
- Option A: [tex]\(3x + 2y = 6\)[/tex]
Solve for [tex]\(y\)[/tex]:
[tex]\[ 2y = -3x + 6 \][/tex]
Divide by 2:
[tex]\[ y = -\frac{3}{2}x + 3 \][/tex]
The slope here is [tex]\(-\frac{3}{2}\)[/tex].
- Option B: [tex]\(3x + 4y = 6\)[/tex]
Solve for [tex]\(y\)[/tex]:
[tex]\[ 4y = -3x + 6 \][/tex]
Divide by 4:
[tex]\[ y = -\frac{3}{4}x + \frac{3}{2} \][/tex]
The slope here is [tex]\(-\frac{3}{4}\)[/tex].
- Option C: [tex]\(2x + 4y = 6\)[/tex]
Solve for [tex]\(y\)[/tex]:
[tex]\[ 4y = -2x + 6 \][/tex]
Divide by 4:
[tex]\[ y = -\frac{1}{2}x + \frac{3}{2} \][/tex]
The slope here is [tex]\(-\frac{1}{2}\)[/tex].
- Option D: [tex]\(2x + 6y = 3\)[/tex]
Solve for [tex]\(y\)[/tex]:
[tex]\[ 6y = -2x + 3 \][/tex]
Divide by 6:
[tex]\[ y = -\frac{1}{3}x + \frac{1}{2} \][/tex]
The slope here is [tex]\(-\frac{1}{3}\)[/tex].
4. Identify the perpendicular line:
The only line whose slope is [tex]\(-\frac{3}{2}\)[/tex], which is the negative reciprocal of [tex]\(\frac{2}{3}\)[/tex], is the line given in Option A.
Therefore, the correct choice is:
Option A: [tex]\(3x + 2y = 6\)[/tex]
1. Find the slope of the given equation:
We start by rewriting [tex]\(-2x + 3y = 6\)[/tex] in the slope-intercept form of [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] represents the slope.
[tex]\[ -2x + 3y = 6 \][/tex]
Add [tex]\(2x\)[/tex] to both sides:
[tex]\[ 3y = 2x + 6 \][/tex]
Divide both sides by 3:
[tex]\[ y = \frac{2}{3}x + 2 \][/tex]
So, the slope [tex]\(m\)[/tex] of the given line is [tex]\(\frac{2}{3}\)[/tex].
2. Determine the slope of the perpendicular line:
Lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex]. Therefore, if one line has a slope [tex]\(m\)[/tex], the perpendicular line will have a slope of [tex]\(-\frac{1}{m}\)[/tex].
For our slope, [tex]\(\frac{2}{3}\)[/tex], the perpendicular slope would be:
[tex]\[ -\frac{1}{\left(\frac{2}{3}\right)} = -\frac{3}{2} \][/tex]
3. Compare the slopes of the given options:
- Option A: [tex]\(3x + 2y = 6\)[/tex]
Solve for [tex]\(y\)[/tex]:
[tex]\[ 2y = -3x + 6 \][/tex]
Divide by 2:
[tex]\[ y = -\frac{3}{2}x + 3 \][/tex]
The slope here is [tex]\(-\frac{3}{2}\)[/tex].
- Option B: [tex]\(3x + 4y = 6\)[/tex]
Solve for [tex]\(y\)[/tex]:
[tex]\[ 4y = -3x + 6 \][/tex]
Divide by 4:
[tex]\[ y = -\frac{3}{4}x + \frac{3}{2} \][/tex]
The slope here is [tex]\(-\frac{3}{4}\)[/tex].
- Option C: [tex]\(2x + 4y = 6\)[/tex]
Solve for [tex]\(y\)[/tex]:
[tex]\[ 4y = -2x + 6 \][/tex]
Divide by 4:
[tex]\[ y = -\frac{1}{2}x + \frac{3}{2} \][/tex]
The slope here is [tex]\(-\frac{1}{2}\)[/tex].
- Option D: [tex]\(2x + 6y = 3\)[/tex]
Solve for [tex]\(y\)[/tex]:
[tex]\[ 6y = -2x + 3 \][/tex]
Divide by 6:
[tex]\[ y = -\frac{1}{3}x + \frac{1}{2} \][/tex]
The slope here is [tex]\(-\frac{1}{3}\)[/tex].
4. Identify the perpendicular line:
The only line whose slope is [tex]\(-\frac{3}{2}\)[/tex], which is the negative reciprocal of [tex]\(\frac{2}{3}\)[/tex], is the line given in Option A.
Therefore, the correct choice is:
Option A: [tex]\(3x + 2y = 6\)[/tex]
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