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Sagot :
To find the equation of a line parallel to line EF and passing through the point [tex]\((0, -2)\)[/tex], let's follow these steps:
1. Identify the Slope of Line EF:
The equation of line EF is given as [tex]\( y = \frac{1}{2} x + 6 \)[/tex]. In the slope-intercept form [tex]\( y = mx + b \)[/tex], [tex]\( m \)[/tex] represents the slope. Therefore, the slope of line EF is [tex]\( \frac{1}{2} \)[/tex].
2. Determine the Slope of the Parallel Line:
Parallel lines have the same slope. Hence, the slope of the line we are looking for is also [tex]\( \frac{1}{2} \)[/tex].
3. Use the Point-Slope Form:
The point-slope form of the line equation 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.
Given point: [tex]\( (0, -2) \)[/tex]
Substituting [tex]\( m = \frac{1}{2} \)[/tex], [tex]\( x_1 = 0 \)[/tex], and [tex]\( y_1 = -2 \)[/tex] into the equation:
[tex]\[ y - (-2) = \frac{1}{2} (x - 0) \][/tex]
4. Simplify the Equation:
[tex]\[ y + 2 = \frac{1}{2} x \][/tex]
To put it in the slope-intercept form [tex]\( y = mx + b \)[/tex], subtract 2 from both sides:
[tex]\[ y = \frac{1}{2} x - 2 \][/tex]
Therefore, the equation of the line parallel to line EF and passing through the point [tex]\((0, -2)\)[/tex] is:
[tex]\[ y = \frac{1}{2} x - 2 \][/tex]
Thus, the correct answer is:
[tex]\[ y = \frac{1}{2} x - 2 \][/tex]
1. Identify the Slope of Line EF:
The equation of line EF is given as [tex]\( y = \frac{1}{2} x + 6 \)[/tex]. In the slope-intercept form [tex]\( y = mx + b \)[/tex], [tex]\( m \)[/tex] represents the slope. Therefore, the slope of line EF is [tex]\( \frac{1}{2} \)[/tex].
2. Determine the Slope of the Parallel Line:
Parallel lines have the same slope. Hence, the slope of the line we are looking for is also [tex]\( \frac{1}{2} \)[/tex].
3. Use the Point-Slope Form:
The point-slope form of the line equation 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.
Given point: [tex]\( (0, -2) \)[/tex]
Substituting [tex]\( m = \frac{1}{2} \)[/tex], [tex]\( x_1 = 0 \)[/tex], and [tex]\( y_1 = -2 \)[/tex] into the equation:
[tex]\[ y - (-2) = \frac{1}{2} (x - 0) \][/tex]
4. Simplify the Equation:
[tex]\[ y + 2 = \frac{1}{2} x \][/tex]
To put it in the slope-intercept form [tex]\( y = mx + b \)[/tex], subtract 2 from both sides:
[tex]\[ y = \frac{1}{2} x - 2 \][/tex]
Therefore, the equation of the line parallel to line EF and passing through the point [tex]\((0, -2)\)[/tex] is:
[tex]\[ y = \frac{1}{2} x - 2 \][/tex]
Thus, the correct answer is:
[tex]\[ y = \frac{1}{2} x - 2 \][/tex]
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