Math

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

+−= xy

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.

)( 11 xxmyy −=− This is the general form of the point-slope equation

( ) ( )[ ]6 3 2

3 −−−=−− xy I plugged in my given slope and ordered pair

( )6 3 2

3 +−=+ xy I evaluated any signs next to each other

)6( 3 2

3 2

3 −−=+ xy I distributed the 3 2

− to each term inside the parentheses

4 3 2

3 −−=+ xy I show here the distribution of the 3 2

− and multiplied

3 2

− times 6, which is –4

34 3 2

−−−= xy I subtracted 3 from both sides, moving like-terms together

so I can combine them

7 3 2

−−= xy 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 14 −−= xy 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.

)( 11 xxmyy −=− This is the general form of the point-slope equation

( )0 4 1

5 −=− xy I plugged in my given slope and ordered pair

( )0 4 1

4 1

5 −=− xy I distributed the 4 1

xy 4 1

5 =− I multiplied ( )0 4 1

−

5 4 1

+= xy 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.]