What is the equation of the line: * parallel to the line y = -¼x + 5 and * passing through the point (2, -1)
y = -¼x + 4
y = ¼x + 2
y = ¼x - 1
y = -¼x - 1/2
you have a point (2,-1) and a slope -1/4, so use the point-slope form:
y+1 = -1/4 (x-2)
now just rearrange things as needed.
thank you!
To find the equation of a line parallel to a given line and passing through a given point, you can follow these steps:
1. Determine the slope of the given line. The equation of the given line, y = -¼x + 5, is in slope-intercept form (y = mx + b), where m represents the slope. In this case, the slope is -¼.
2. Use the slope from step 1 and the point (2, -1) to write the equation of the new line. Substitute the coordinates of the point (x, y) into the point-slope form of a linear equation, y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) are the coordinates of the given point. In this case, the coordinates of the point are (2, -1), so the equation becomes: y -(-1) = -¼(x - 2).
3. Simplify the equation found in step 2 to obtain the final equation in slope-intercept form (y = mx + b). Distribute the -¼ on the right-hand side: y + 1 = -¼x + ½. Then, move the constant term to the left side: y = -¼x + ½ - 1. Simplifying further, we have: y = -¼x - ½.
Therefore, the equation of the line parallel to y = -¼x + 5 and passing through the point (2, -1) is y = -¼x - ½.