Week One Assignement

INSTRUCTOR GUIDANCE EXAMPLE: Week Three Discussion Parallel and Perpendicular For this week’s discussion I am going to find the equations of lines that are parallel or perpendicular to the given lines and which are passing through the specified point. First I will work on the equation for the parallel line. The equation I am given is The parallel line must pass through point y = -⅔ x + 2 (-6, -3)

I have learned that a line parallel to another line has the same slope as the other line, so now I know that the slope of my parallel line will be -⅔. Since I now have both the slope and an ordered pair on the line, I am going to use the point-slope form of a linear equation to write my new equation. y – y1 = m(x – x1) y – (-3) = -⅔[x – (-6)] y + 3 = -⅔x + (-⅔)6 y = -⅔x – 4 – 3 y = -⅔x – 7 This is the general form of the point-slope equation Substituting in my known slope and ordered pair Simplifying double negatives and distributing the slope Because (-⅔)6 = -4 and 3 is subtracted from both sides The equation of my parallel line!

This line falls as you go from left to right across the graph of it, the y-intercept is 7 units below the origin, and the x-intercept is 10.5 units to the left of the origin. Now I am ready to write the equation of the perpendicular line. The equation I am given is y = -4x – 1 The perpendicular line must pass through point (0, 5) I have learned that a line perpendicular to another line has a slope which is the negative reciprocal of the slope of the other line so the first thing I must do is find the negative reciprocal of –4. The reciprocal of -4 is -¼ , and the negative of that is –(-¼) = ¼. Now I know my slope is ¼ and my given point is (0, 5). Again I will use the point-slope form of a linear equation to write my new equation. y – y1 = m(x – x1) y – 5 = ¼ (x – 0) y–5=¼x y=¼x+5

Substituting in the slope and ordered pair The zero term disappears Adding 5 to both sides of the equation The equation of my perpendicular line!

This line...