Upcoming SlideShare
×

# 10 Perpendicular Lines

2,116 views

Published on

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
2,116
On SlideShare
0
From Embeds
0
Number of Embeds
38
Actions
Shares
0
0
0
Likes
0
Embeds 0
No embeds

No notes for slide

### 10 Perpendicular Lines

1. 1. Perpendicular Lines 10.
2. 2. A (4, 4) D (-4, 4) B (4, -4) C (-4, -4) 90 0 90 0
3. 3. Perpendicular Lines 0 y x A (a , b) B (-b , a) After a 90 0 rotation (anti-clockwise) A moves to B As both lines go through origin gradients are: 90 0 NB Fraction has been turned upside down and turned into negative
4. 4. Copy the following: For perpendicular lines: 1. Lines are perpendicular 2. To find gradient of line perpendicular to given line think: “ Flip the fraction and change the sign”
5. 5. Example 1 What is the gradient of a line perpendicular to the line through the points P (4 , 7) and Q (5 , 11) ? Solution : <ul><li>Find gradient between </li></ul><ul><li>given points </li></ul>Flip fraction- change sign 2. Use m 1 x m 2 = -1 to find gradient of perpendicular line.
6. 6. Example 2 Two lines have equations 3x + 2y - 7 = 0 & 4x - 6y + 1 = 0 Prove that they are perpendicular. Solution : <ul><li>Find the gradient of line 1 </li></ul><ul><li>by rearranging and using </li></ul><ul><li>y = mx + c </li></ul>Line 1 3x + 2y - 7 = 0 2y = -3x + 7 Gradient 1 = -3/2 Line 2 4x - 6y + 1 = 0 6y = 4x + 1 Gradient 2 = 2/3 2. Find the gradient of line 2 by rearranging and using y = mx + c
7. 7. Example 2 Two lines have equations 3x + 2y - 7 = 0 & 4x - 6y + 1 = 0 Prove that they are perpendicular. Solution : 3. Check that m 1 x m 2 = -1 4. Make statement Since m 1 x m 2 = -1 lines are perpendicular
8. 8. Heinemann, p.6, Ex 1D, Q1, 2, 3 & 6