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how do you work out gradient?

if you have got a agra[h with a sytraight line on it, how do you work out teh gradient, i think know the formula m=(y1-y2) / (x1- x2) but am struggling to use it, can someone give me an example please!?

thanks!
Two points on the straight line: say, (6,4) and (-3,2)

We'll assume that (6,4) is (x1,y1) and (-3,2) is (x2,y2)

Use the formula: m=y2-y1/x2-x1
=2-4/-3-6
=-2/-9

So the gradient of the straight line is 2/9 (two negatives equal a positive).
Reply 2
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786_786
OP
1
Ramadulla
Two points on the straight line: say, (6,4) and (-3,2)

We'll assume that (6,4) is (x1,y1) and (-3,2) is (x2,y2)

Use the formula: m=y2-y1/x2-x1
=2-4/-3-6
=-2/-9

So the gradient of the straight line is 2/9 (two negatives equal a positive).


i know im bugging now but is there anyyyy chance you can give me an example and see if i can work it out please!

thank you soo much!
Ramadulla
Two points on the straight line: say, (6,4) and (-3,2)

We'll assume that (6,4) is (x1,y1) and (-3,2) is (x2,y2)

Use the formula: m=y2-y1/x2-x1
=2-4/-3-6
=-2/-9

So the gradient of the straight line is 2/9 (two negatives equal a positive).


Couldn't of explained it better, myself!
Reply 4
Avatar for nuodai
nuodai
17
786_786
i know im bugging now but is there anyyyy chance you can give me an example and see if i can work it out please!

thank you soo much!

He... just did.
Reply 5
Avatar for 786_786
786_786
OP
1
nuodai
He... just did.

yeh but the answers there so i dont know if id be able to work it out in the exam myself...ah never mind ill find an example on th net somewhere
Reply 6
Avatar for Kyalimers
Kyalimers
21
786_786
i know im bugging now but is there anyyyy chance you can give me an example and see if i can work it out please!

thank you soo much!

Work out the gradient of a line which passes through the x-axis at 4, and which also passes through the point (14,9)
786_786
i know im bugging now but is there anyyyy chance you can give me an example and see if i can work it out please!

thank you soo much!


Example:
Work out the gradient of line A and line B where A(10,9) and B(6,-7). Make an equation in the form of y=mx+c

Working:
Okay, so we have coordinates of A and B, so we use the formula: m=(y2-y1)/(x2-x2). So substitute the coordinates of both A and B into the formula: m=(-7-9)/(6-10), which then equals: m=-16/-4 which simplifies to +4.

We then substitute into the formula, and can use any of the points above for either A and B. So we substitute our gradient (m) into the (m) of the formula and any point, we will pick point A (easier as no negative or positive signs-less likely to have a mistake).

So: y-9=4(x-10)
y-9=4x+40
y=4x+49

That's and example! :yep:
Reply 8
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786_786
OP
1
sohanshah
Work out the gradient of a line which passes through the x-axis at 4, and which also passes through the point (14,9)

is it 0.8? :o:
Reply 9
Avatar for Kyalimers
Kyalimers
21
9/10

remembering the first co-ordinate is (4,0)
Reply 10
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786_786
OP
1
sohanshah
9/10

remembering the first co-ordinate is (4,0)


:s-smilie: forget it ill just miss the gradient question out!
786_786
:s-smilie: forget it ill just miss the gradient question out!

what exam is this for?
Reply 12
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minesapint
It's worth pursuing this as it's money for old rope once you've got it.
if you have A(x1,y1),B(x2,y2)A(x_1,y_1), B(x_2,y_2)

then the gradient between the two points is y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1}
Reply 14
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Alvrae
Original post by 786_786
i know im bugging now but is there anyyyy chance you can give me an example and see if i can work it out please!

thank you soo much!


Level A = 16.050
Level B = 16.042

Distance between = 2.4m
change in level = 0.008

0.008/2.4 = Gradient
2.4/0.008 = 1 in 300

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