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# The Gradient Formula - PowerPoint PPT Presentation

The Gradient Formula. Gradient of AB = m AB = height / horizontal = BC / AC = (y 2 -y 1 ) (x 2 -x 1 ). B(x 2 ,y 2 ). y 2 -y 1. A(x 1 ,y 1 ). x 2 -x 1. C(x 2 ,y 1 ). The Gradient Formula ctd. Summary If A is (x 1 ,y 1 ) and B is (x 2 ,y 2 )

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Presentation Transcript
• = mAB
• = height/horizontal
• = BC/AC
• = (y2-y1) (x2-x1)

B(x2,y2)

y2-y1

A(x1,y1)

x2-x1

C(x2,y1)

• Summary
• If A is (x1,y1) and B is (x2,y2)
• then the gradient of AB is given by
• mAB = (y2-y1) (x2-x1)
Example 4

P is (5, 9) and Q is (7,17).

• mPQ = (y2-y1) (x2-x1)
• = (17 - 9)
• ( 7 - 5)
• = 8/2
• = 4

NB: line looks like

Q

P

Example 5

E is (15, 9) and F is (7,17).

• mEF = (y2-y1) (x2-x1)
• = (17 - 9)
• ( 7 - 15)
• = 8/-8
• = -1

NB: line looks like

F

E

Example 6

V is (-5, -9) and W is (12,-9).

• mVW = (y2-y1) (x2-x1)
• = (-9 -(- 9))
• ( 12 - (-5))
• = 0/17
• = 0
• NB: A horizontal line has no steepness.

NB: line looks like

V

W

Example 7

S is (5, -9) and T is (5, 12).

• mST = (y2-y1) (x2-x1)
• = (12 -(- 9))
• ( 5 - 5)
• = 21/0
• and this is undefined
• or infinite

NB: line looks like

T

S

Parallel Lines
• Parallel lines run in the same direction so must be equally steep.
• Hence parallel lines have equal gradients.
• Example 8
• Prove that if A is (4,-3) , B is (9,3) C is (11,1) & D is (2, -1)
• then ACBD is a parallelogram
Ex8 ctd

D

B

NB; The order of the letters is important.

A

• mAC = (1 + 3)/(11- 4) = 4/7
• mDB = (3 + 1)/(9 - 2) = 4/7
• mAC = mDB so AC is parallel to DB
• mAD = (-1 + 3)/(2 - 4) = 2/-2 = -1
• mCB = (3 - 1)/(9 - 11) = 2/-2 = -1
• mAD = mCB so AC is parallel to DB
• Since the opposite sides are parallel then it follows that ACBD is a parallelogram.

C

COLLINEARITY

Defn: Three or more points are said to be collinear if the gradients from any one point to all the others is always the same.

Example 8a

K is (5, -8), L is (-2, 6) and M is (9, -16). Prove that the three points are collinear.

6 - (-8)

14

= -2

mKL =

=

Since KL & KM have equal gradients and a common point K then it follows that K, L & M are collinear.

-2 - 5

-7

-16 - (-8)

-8

mKM =

=

= -2

9 - 5

4

Ex8b

A Navy jet flies over two lighthouses with map coordinates (210,115) & (50,35). If it continues on the same path will it pass over a yacht at (10,15) ?

m1 = (115-35)/(210-50)

= 80/160

= 1/2

m2 = (115-15)/(210-10)

= 100/200

= 1/2

Since gradients equal & (210,115) a common point then the three places are collinear so plane must fly over all three.