1. Day 27 – Slope criteria of special lines

2. Introduction
Two lines which intersect at a right angle are said to be perpendicular. If two or more lines on the same plane do not meet, they are said to be parallel. The slopes of parallel are equal while the product of that of perpendicular lines is a negative one. In this lesson, we will prove these properties.

3. Vocabulary Slope of a Line
This is a number that measures steepness of a line.

4. If two lines are parallel then they have the same slope
If two lines are parallel then they have the same slope. Consider the figure below.

5. Line MN has translated zero units horizontally and 4 units vertically downwards to line 𝑀 ′ 𝑁 ′ . If 𝑎 represents one unit, then point P 𝑥 1, 𝑦 1 becomes 𝑃 ′ 𝑥 1, 𝑦 1 −4𝑎 and point Q 𝑥 2 , 𝑦 2 becomes 𝑄 ′ 𝑥 2 , 𝑦 2 −4𝑎 since line MN is translated 4 units downwards. We show that Lines MN and 𝑀 ′ 𝑁 ′ have equal slopes.

6. Proof
We want to prove that the slope of MN is equal to the slope of 𝑀 ′ 𝑁 ′ .
Slope of MN = ∆𝑦 ∆𝑥
= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1
Slope of 𝑀 ′ 𝑁 ′ = ∆𝑦 ∆𝑥
= 𝑦 2 −4𝑎− 𝑦 1 −4𝑎 𝑥 2 − 𝑥 1 = 𝑦 2 −4𝑎+ 𝑦 1 +4𝑎 𝑥 2 − 𝑥 1
Thus, Slope of MN = Slope of 𝑀 ′ 𝑁 ′

7. Example 1
Line AB is parallel to CD. Line AB passes through points R(4,9) and S(2,3) while CD passes through C(0,11) and D(-2,5). Compare the slopes of CD and AB?
Solution
Slope of AB = 9−3 4−2 = = 3
Slope of CD = 5−11 −2−0 = = 3
They have the same slope.
The product of the slopes of two perpendicular lines is -1.
Consider the figure below. One unit represents a.

9. Two perpendicular lines KL and ST intersect at point V
Two perpendicular lines KL and ST intersect at point V. The product of their slopes is -1.
Proof
Slope of ST = 𝑦 1 +2𝑎− 𝑦 1 𝑥 1 +2𝑎− 𝑥 1 = 2𝑎 2𝑎 = 1
Slope of KL = 𝑦 1 +2𝑎− 𝑦 1 𝑥 1 +2𝑎− 𝑥 1 +4𝑎 = 2𝑎 −2𝑎 = -1
Slope of ST × Slope of KL = 1 × -1 = -1
Thus the product of the two lines is -1.

10. Example 2.
Line AB passes through points A(4,2) and B(2,1). If another line ST is perpendicular to AB, and passes through (-2,4) and (-3,6) what is the slope of AB and ST? What is the product of their slope?
Solution
Slope of AB = ∆𝑦 ∆𝑥
= 2−1 4−2 = 1 2
Slope of ST = ∆𝑦 ∆𝑥
= 6−4 −3−−2 = 2 −1 =−2
The product is −2× 1 2 =−1.

11. homework
Lines HL and JK are parallel. Line HL passes through points H(3,4) and L(7,9). What is the slope of line JK? If line HL is perpendicular to line MN and passes through M(7,1) and N(12,-3) , What is the slope of line MN? What is the product of the slopes of lines MN and JK?

12. Answers to homework Slope of line JK = 5 4 Slope of line MN = −4 5
The product is -1.

13. THE END