## Ashford University Math 221 Week 3 Assignment Part

Our assignment is to find the equations of the lines that are parallel or perpendicular to the lines that were assigned to us. My assigned number was 26. The equation given to me was y=−12x+1The parallel line must pass through the ordered pair(-2, 3)When two lines are parallel, they have to have the same slope. So the point-slope equation is y-y1 = m(x-x1)After plugging in the formula, I got y-3x−(−2)¿−12¿This can then be simplified to equal y-3= ¿−12(x+2)After using the distributive property of multiplication, we get y−3=−12x+2Then I’m going to add 3 to both sides and I’d get Y=−12x+5Since the slope of −12is negative, then the line will go from left to right. The equation of the parallel line is Y=−12x+5.Starting from the origin, the 5 will be plotted five spaces to the left at the y-intercept.For the second part of the discussion, I will write the equation of the perpendicular line.

Being that my first name starts with “J” The equation which will be used by me is y = -x+ 2 and the parallel line must pass through ⅔ point (9, -3). The slope of the given line is -. So, the ⅔ slope of the parallel line will also be -. We now ⅔ know the slope of the new line and we know that the line passes through (9,-3). I will therefore use the point-slope form of a linear equation to write my new equation. y – y1 = m(x – x1) This is the general form of the point-slope equation y – (-3) = -x [x – (9)] ⅔ Substituting in my known slope and ordered pair y + 3 = -x + (-)-9 ⅔ ⅔ Simplifying double negatives and distributing the slope y = -x +6 – 3 ⅔-3 is subtracted from both sides y = -x + 3 ⅔ Equation of my parallel line! This line falls as one goes from left to right across the graph of it, the y-intercept is -3 units below the origin , and the x-intercept is 4.5 unit to the left of the origin . Now we have to write the equation of the perpendicular line.

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