Answer :
To find the equation of the line that is parallel to the given line and passes through the point [tex]\((-2, 2)\)[/tex], we must follow these steps:
1. Identify the slope of the given line. The general form of the line equation [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope. Looking at the given equation [tex]\(y = \frac{1}{5} x + 4\)[/tex], we see that the slope [tex]\(m\)[/tex] is [tex]\(\frac{1}{5}\)[/tex].
2. Recognize that a line parallel to the given line will have the same slope. Therefore, the slope of our new line will also be [tex]\(\frac{1}{5}\)[/tex].
3. Next, we need to find the y-intercept [tex]\(b\)[/tex] for our new line. We have the slope [tex]\(m = \frac{1}{5}\)[/tex] and a point [tex]\((-2, 2)\)[/tex] through which the line passes.
4. Substitute the point [tex]\((-2, 2)\)[/tex] into the equation [tex]\(y = mx + b\)[/tex] to find the y-intercept [tex]\(b\)[/tex]:
[tex]\[ 2 = \frac{1}{5}(-2) + b \][/tex]
5. Solve for [tex]\(b\)[/tex]:
[tex]\[ 2 = -\frac{2}{5} + b \][/tex]
To isolate [tex]\(b\)[/tex], add [tex]\(\frac{2}{5}\)[/tex] to both sides:
[tex]\[ 2 + \frac{2}{5} = b \][/tex]
6. Convert [tex]\(2\)[/tex] to a fraction with a common denominator to facilitate addition:
[tex]\[ 2 = \frac{10}{5} \][/tex]
Thus,
[tex]\[ \frac{10}{5} + \frac{2}{5} = b \][/tex]
7. Add the fractions:
[tex]\[ b = \frac{12}{5} \][/tex]
8. Substitute the slope [tex]\(m = \frac{1}{5}\)[/tex] and the y-intercept [tex]\(b = \frac{12}{5}\)[/tex] back into the slope-intercept form of the line equation:
[tex]\[ y = \frac{1}{5} x + \frac{12}{5} \][/tex]
Therefore, the equation of the line that is parallel to the given line and passes through the point [tex]\((-2, 2)\)[/tex] is:
[tex]\[ y = \frac{1}{5} x + \frac{12}{5} \][/tex]
1. Identify the slope of the given line. The general form of the line equation [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope. Looking at the given equation [tex]\(y = \frac{1}{5} x + 4\)[/tex], we see that the slope [tex]\(m\)[/tex] is [tex]\(\frac{1}{5}\)[/tex].
2. Recognize that a line parallel to the given line will have the same slope. Therefore, the slope of our new line will also be [tex]\(\frac{1}{5}\)[/tex].
3. Next, we need to find the y-intercept [tex]\(b\)[/tex] for our new line. We have the slope [tex]\(m = \frac{1}{5}\)[/tex] and a point [tex]\((-2, 2)\)[/tex] through which the line passes.
4. Substitute the point [tex]\((-2, 2)\)[/tex] into the equation [tex]\(y = mx + b\)[/tex] to find the y-intercept [tex]\(b\)[/tex]:
[tex]\[ 2 = \frac{1}{5}(-2) + b \][/tex]
5. Solve for [tex]\(b\)[/tex]:
[tex]\[ 2 = -\frac{2}{5} + b \][/tex]
To isolate [tex]\(b\)[/tex], add [tex]\(\frac{2}{5}\)[/tex] to both sides:
[tex]\[ 2 + \frac{2}{5} = b \][/tex]
6. Convert [tex]\(2\)[/tex] to a fraction with a common denominator to facilitate addition:
[tex]\[ 2 = \frac{10}{5} \][/tex]
Thus,
[tex]\[ \frac{10}{5} + \frac{2}{5} = b \][/tex]
7. Add the fractions:
[tex]\[ b = \frac{12}{5} \][/tex]
8. Substitute the slope [tex]\(m = \frac{1}{5}\)[/tex] and the y-intercept [tex]\(b = \frac{12}{5}\)[/tex] back into the slope-intercept form of the line equation:
[tex]\[ y = \frac{1}{5} x + \frac{12}{5} \][/tex]
Therefore, the equation of the line that is parallel to the given line and passes through the point [tex]\((-2, 2)\)[/tex] is:
[tex]\[ y = \frac{1}{5} x + \frac{12}{5} \][/tex]
