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# how do you work out gradient?

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1. 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!
2. 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).
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!
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!
5. (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!
He... just did.
6. (Original post by 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
7. (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!
Work out the gradient of a line which passes through the x-axis at 4, and which also passes through the point (14,9)
8. (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!
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!
9. (Original post by 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?
10. 9/10

remembering the first co-ordinate is (4,0)
11. (Original post by sohanshah)
9/10

remembering the first co-ordinate is (4,0)
forget it ill just miss the gradient question out!
12. (Original post by 786_786)
forget it ill just miss the gradient question out!
what exam is this for?
13. It's worth pursuing this as it's money for old rope once you've got it.
14. if you have

then the gradient between the two points is
15. (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

2.4/0.008 = 1 in 300

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