Read the following instructions in order to complete this discussion, and review the example of how to complete the math required for this assignment:Given an equation of a line, find equations for lines parallel or perpendicular to it going through specified points. Find the appropriate equations and points from the table below. Simplify your equations into slope-intercept form.Use your assigned number to complete.Discuss the steps necessary to carry out each activity. Describe briefly what each line looks like in relation to the original given line.Answer these two questions briefly in your own words: What does it mean for one line to be parallel to another?What does it mean for one line to be perpendicular to another?Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing your math work.):Origin Ordered pair X- or y-intercept Slope ReciprocalYour initial post should be 150-250 words in length.The equation is # 13

Number 13

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 2 3 2 + − = x y The parallel line must pass through point ( )3 ,6 − − = = 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 3 2 −. 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. ) ( 1 1 x x m y y − = − = = =This is the general form of the point -slope equation= ( ) ( ) [ ] 6 3 2 3 − − − = − − x y I plugged in my given slope and ordered pair ( )6 3 2 3 + − = + x y I evaluated any signs next to each other )6( 3 2 3 2 3 − − = + x y I distributed the 32 −=to each term inside the parentheses = 4 3 2 3 − − = + x y I show here the distribution of the 3 2 − =and multiplied== = ==== 3 2 − =times 6, which is 4= = 3 4 3 2 − − − = x y I subtracted 3 from both sides, moving like -terms together so I can combine them 7 3 2 − − = x y Like-terms are combined, and the result is 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 will write the equation of the perpendicular line. The equation I am given is 1 4− − = x y The perpendicular line must pass through point () 5 , 0 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 4 1 −, and the negative of that is 4 1 4 1 = − − . Now I know my slope is 4 1 and my given point is = () 5 , 0 . Again, I will use the point -slope form of a linear equation to write my new equation. ) ( 1 1 x x m y y − = − = = =This is the general form of the point -slope equation= ( ) 0 4 1 5 − = − x y I plugged in my given slope and ordered pair ( )0 4 1 4 1 5 − = − x y I distributed the 4 1 x y 4 1 5 = −= = ==I multiplied= ( )0 4 1 − = = 5 4 1 + = x y I add 5 to both sides of the equation, and the result is the equation of my perpendicular line! This line rises as you move from left to right across the graph. The y -intercept is five units above the original and the x -intercept is 20 units to the left of the origin. [The answers to part d of the discussion will vary with students’ understanding.]

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