Answer :
To find the equation of the line that passes through the point [tex]\((0, 1)\)[/tex] and is perpendicular to the line [tex]\(y = -2x + 2\)[/tex], follow these steps:
1. Find the slope of the given line [tex]\(y = -2x + 2\)[/tex]:
The equation is in slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope.
Here, the slope [tex]\(m\)[/tex] of the line is [tex]\(-2\)[/tex].
2. Determine the slope of the line perpendicular to the given line:
The slope of a line perpendicular to another is the negative reciprocal of the slope of the given line.
The negative reciprocal of [tex]\(-2\)[/tex] is [tex]\(\frac{1}{2}\)[/tex].
3. Use the point-slope form of the equation of a line to find the equation of the required line:
The point-slope form is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\((x_1, y_1)\)[/tex] is a point on the line and [tex]\(m\)[/tex] is the slope.
Here, we have the point [tex]\((0, 1)\)[/tex], and the slope is [tex]\(\frac{1}{2}\)[/tex].
4. Substitute the values into the point-slope form:
[tex]\[ y - 1 = \frac{1}{2}(x - 0) \][/tex]
5. Simplify to obtain the equation in slope-intercept form [tex]\(y = mx + b\)[/tex]:
[tex]\[ y - 1 = \frac{1}{2}x \][/tex]
Add 1 to both sides to isolate [tex]\(y\)[/tex]:
[tex]\[ y = \frac{1}{2}x + 1 \][/tex]
Therefore, the equation of the line that passes through [tex]\((0, 1)\)[/tex] and is perpendicular to the line [tex]\(y = -2x + 2\)[/tex] is:
[tex]\[ y = \frac{1}{2}x + 1 \][/tex]
1. Find the slope of the given line [tex]\(y = -2x + 2\)[/tex]:
The equation is in slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope.
Here, the slope [tex]\(m\)[/tex] of the line is [tex]\(-2\)[/tex].
2. Determine the slope of the line perpendicular to the given line:
The slope of a line perpendicular to another is the negative reciprocal of the slope of the given line.
The negative reciprocal of [tex]\(-2\)[/tex] is [tex]\(\frac{1}{2}\)[/tex].
3. Use the point-slope form of the equation of a line to find the equation of the required line:
The point-slope form is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\((x_1, y_1)\)[/tex] is a point on the line and [tex]\(m\)[/tex] is the slope.
Here, we have the point [tex]\((0, 1)\)[/tex], and the slope is [tex]\(\frac{1}{2}\)[/tex].
4. Substitute the values into the point-slope form:
[tex]\[ y - 1 = \frac{1}{2}(x - 0) \][/tex]
5. Simplify to obtain the equation in slope-intercept form [tex]\(y = mx + b\)[/tex]:
[tex]\[ y - 1 = \frac{1}{2}x \][/tex]
Add 1 to both sides to isolate [tex]\(y\)[/tex]:
[tex]\[ y = \frac{1}{2}x + 1 \][/tex]
Therefore, the equation of the line that passes through [tex]\((0, 1)\)[/tex] and is perpendicular to the line [tex]\(y = -2x + 2\)[/tex] is:
[tex]\[ y = \frac{1}{2}x + 1 \][/tex]
