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The Gradient Formula

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- Gradient of AB
- = 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)

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

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

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

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

Vertical line has infinite gradient.

S

Positive gradient goes uphill.

Negative gradient goes downhill.

Zero gradient is horizontal.

Infinite gradient is vertical.

- 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

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.