GCSE maths question

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Caporegime
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I seem to be struggling again. I'm tired and overthinking this, every time I look at this I get a different answer depending on where I put the numbers in the formula, so I turn to you GD, put me out of my misery!

Find the gradient of the line that passes through points (-4, 3) and (2, -1).

So far I've come up with 1.5, -1.5 and 0.666

Help!

Cheers :)
 
I have literally 0% clue what the **** this question is going on about.

Interesting how much the maths syllabus has changed since i sat GCSE because i did nothing of this sort when i was in school. Though it is national pi day so happy pi day.
 
harry5522 said:
I have literally 0% clue what the **** this question is going on about.

Interesting how much the maths syllabus has changed since i sat GCSE because i did nothing of this sort when i was in school. Though it is national pi day so happy pi day.
Click to expand...

You really should have. Gradient of a straight line is pretty standard in the GCSE syllabus and has been for as long as GCSEs have existed. y = mx + c ring any bells?
 
Exactly. We're going back almost 20 years since I did this, and I haven't had time to do it properly what with working 7 days a week almost every single week. FFS.

At least learning this time around I'm doing my test the night before it's due. Consistency is key I suppose :D


Thanks for the help guys, much appreciated.
 
Mason- said:
You really should have. Gradient of a straight line is pretty standard in the GCSE syllabus and has been for as long as GCSEs have existed. y = mx + c ring any bells?
Click to expand...

I don't recall doing gradient of a straight line at all boss. Wish i had, the more knowledge the better. A lot of the things we did get taught are now however quite lost on me as they are not in daily use situations. I imagine some WD-40 and some practice questions and it would start to come back to me however i think if i had no practice and was make to sit the test now i would probably barley pass or even more likely fail.
 
Change in y / change in x

or y2 - y1 / x2 - x1

so (-1 -3) / (2 - (-4))

so the gradient is negative 0.66 / -0.66
 
Right, let's see how many of you still know this stuff :D


I've come to an answer and so has a colleague and we're both saying the other is wrong, so let's see if GD can resolve this!


Go on, give it your best shot.

IPTHuMj.png
 
Gradient = -(2/3)
Eqn: y - y1 = m(x - x1)
Use first point: y - 3 = -(2/3)(x - (-4))
y - 3 = -(2/3)(x + 4)
y - 3 = -(2x/3) -(8/3)
y = -(2x/3) + (1/3)

Plug first point (-4,3) back in to test: y = (8/3) + (1/3) = 3
Plug second point (2,-1) back in to test: y = -(4/3) + (1/3) = -1

Try new point (3,2): y = (6/3) + (1/3) = 7/3 -> not on line

X intercept occurs at y = 0: -(2x/3) + (1/3) = 0
(2x/3) = (1/3)
2x = 1
x = (1/2)

Y intercept occurs when x = 0: y = (1/3)

grph_zps4iut8ab6.png
 
I missed Touch's answer, and I haven't got the results back yet. I'm just trying to see what the outcome will be as this one had me scratching my head for bloody ages :D


I've got a beer riding on this too!
 
Okay i like these sort of threads. To keep people busy while OP gets the answer, i have decided to make a quadratic to continue on the subject. To stop cheaters using online quadratic solvers, i made one with complex numbers but with the answers being non complex/imaginary.


(x+i-2)(x+i-2) = y

Where i is the base imaginary number. i^2 = -1

Find the values of x when they cross the line i(2x-4)

Edit: edited so the answer is a whole number to make things neater
 
Last edited:
(x+i)^2 + -2(x+i) + -2(x+i) + 4 = 2ix - 4i
(x+i)^2 -4(x+i) + 4 = 2ix - 4i
x^2 + 2ix + i^2 -4x - 4i + 4 = 2ix - 4i
x^2 + i^2 + 4 = 0
x^2 +3 = 0
x^2 = -3

x = +1.7 x = -1.7

where did I go wrong? :p
 
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