Math 221 Intro to Algebra Week 3 DQ 1

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.]