ACT Study Modules Topic: Finding from and perpendicular lines

Khan Academy Link: https://www.khanacademy.org/math/geometry-home/analytic-geometry- topic/parallel-perpendicular-lines-coordinate-/v/classify-lines

To determine the slope of parallel lines from an :

Step one: arrange the equation in the form, = + Step two: determine the slope ( value) Step three: compare the 𝑦𝑦 π‘šπ‘šπ‘šπ‘š 𝑏𝑏 Step four: compare the β€œ ” valuesπ‘šπ‘š

If the b values are different𝑏𝑏 and the slope is the same, then the lines are parallel.

Example: 3 + = 9 5 = 3 Add 3 to both sides add 5 to both sides βˆ’ π‘₯π‘₯ of 𝑦𝑦the equationβˆ’ of𝑦𝑦 theβˆ’ equationπ‘₯π‘₯ π‘₯π‘₯= 3 9 = 3 + 5

In the functions, 𝑦𝑦= 3 π‘₯π‘₯ βˆ’9 & = 3 + 5 the slope is 3, and the𝑦𝑦 valuesπ‘₯π‘₯ are different, making the lines parallel. 𝑦𝑦 π‘₯π‘₯ βˆ’ 𝑦𝑦 π‘₯π‘₯ 𝑏𝑏 To determine the slope of perpendicular lines from an equation:

Step one: arrange the equation in the form = + Step two: determine the slope ( value) Step three: compare the slopes 𝑦𝑦 π‘šπ‘šπ‘šπ‘š 𝑏𝑏 π‘šπ‘š If the slopes are negative reciprocals, the lines are perpendicular.

Example: 5 = 4 + 8 = 1/5 Add 5 to both sides Subtract 8 from both sides 𝑦𝑦 βˆ’ ofπ‘₯π‘₯ the βˆ’equation 𝑦𝑦of the equationβˆ’ π‘₯π‘₯ π‘₯π‘₯= 5 4 = 3 + 5

𝑦𝑦 π‘₯π‘₯ βˆ’ 𝑦𝑦 π‘₯π‘₯ Slopes are 5 and -1/5, the slopes are negative reciprocals, making the lines perpendicular.

Example: In the standard ( , ) coordinate plane which of the lines are perpendicular to the = 2 + 2 that passes through the (0,3) π‘₯π‘₯ 𝑦𝑦 Step𝑦𝑦 one:βˆ’ determineπ‘₯π‘₯ the slope of the perpendicular line:. The slope of the original line is 2 the negative reciprocal of 2 is 1 βˆ’ Step two: = + βˆ’ 2 Step three: to determine1 , substitute the and - values of the point, (0,3) 𝑦𝑦 2 π‘₯π‘₯ 𝑏𝑏 3 = (0) + 𝑏𝑏 π‘₯π‘₯ 𝑦𝑦 3 = 1 2 𝑏𝑏 Step four: write the equation, = + 3 𝑏𝑏 1 𝑦𝑦 2 π‘₯π‘₯ Practice:

Line Equation of a parallel line Equation of a perpendicular line

= 2 5

𝑦𝑦 π‘₯π‘₯ βˆ’

2 3 = 12

π‘₯π‘₯ βˆ’ 𝑦𝑦

3 = 1 4

𝑦𝑦 βˆ’ π‘₯π‘₯ βˆ’

5 + 2 = 15

π‘₯π‘₯ 𝑦𝑦