Maths help

[Brigadier]

Having not done anything past a bit of algebra in thirty years, I'm stumped.

(I can post many more from this particular evening's "fun", from my first-born's homework)

 

[Brigadier]

I didn't get any further than y=mx - c

On the basis that the gradient is positive, and the tangent crosses the x axis in the negative......

 

[Ihavenojob]

Try here for some answers
https://www.funnyexam.com/

 

[SirGalahad]

https://www.bbc.com/bitesize/guides/zqjqsrd/revision/2

I think you will be able to work out the answer - it's straight forward once you get a bit of a nudge.

 

[Ian H]

You’ve lost me.

Is that A-level or Degree stuff?

 

[sodthisforfun]

You’ve lost me.

Is that A-level or Degree stuff?

Primary

(just kidding, I hope?!)

 

[EFLImpudence]

I didn't understand the question.

 

[SirGalahad]

I didn't understand the question.

Nor did Brexiteers.

 

[SirGalahad]

You’ve lost me.

Is that A-level or Degree stuff?

You must be joking. I would say high school or we are in trouble.

 

[Ian H]

You must be joking. I would say high school or we are in trouble.

Yes, just kidding, it looks like the stuff my younger ones get. I’ll ask them if they can do it in the morning

 

[rsgaz]

y = mx + c

The gradient of line PO (not shown) is 4 to 1 'downhill'! As they are perpendicular, the gradient of the line we want is 1 to 4 rising. Therefore 'm' is ¼ or 0.25. So...

y = 0.25x + c

Any point on that line must satisfy the solution, and we are given the point (-1,4), so swap x for -1 and y for 4.

4 = 0.25(-1) + c

Solve for c

4 = -0.25 + c
4.25 = c
c = 4.25

So, the answer is...

y = 0.25x + 4.25

 

[Ian H]

What do you work as rsgaz? You seem to have a great understanding of everything!!

 

[rsgaz]

You can use any online graphing tool to check your answer ( https://www.desmos.com/calculator ).



It crosses the point (-1,4). I added the circle 'manually' to show it fits, because we know the radius is root 17.

 

[Aquaheat]

y = mx + c

The gradient of line PO (not shown) is 4 to 1 'downhill'! As they are perpendicular, the gradient of the line we want is 1 to 4 rising. Therefore 'm' is ¼ or 0.25. So...

y = 0.25x + c

Any point on that line must satisfy the solution, and we are given the point (-1,4), so swap x for -1 and y for 4.

4 = 0.25(-1) + c

Solve for c

4 = -0.25 + c
4.25 = c
c = 4.25

So, the answer is...

y = 0.25x + 4.25

It's been years since I did any of this plus I'm slightly inhebriated but bear with me as it's genuine questions and it may help others too...

I get the coordinates for x and y (-1 and 4 respectively) what I don't get or can't remember is what m and c represent? There is no reference to them on the diagram. Is m the mean? Median?

Also, why does the initial question focus on the P(-1,4) with O being centre? Is O being centre not important to solving it as you don't seem to have included that in your solution? Or is O the start point if the grid reference?

Sorry if I come across a bit dense on this but I'm normally quite good with msths but this has thrown me a bit of a curve ball unless in really over complicating things?

*maths! But apparently english is a bridge too far!

Jon

 

[rsgaz]

what I don't get or can't remember is what m and c represent?

m represents the gradient of the line. Try a few different numbers on the graphing website I linked to, positive gradient gives an 'uphill' line, negative gives a downhill line. See how all lines go through point (0,0)



Which brings us nicely to what c represents. Changing the value for c just moves the line up and down. It's value relates to where the line intersects the y axis.

Is O being centre not important to solving it as you don't seem to have included that in your solution?

I did mention the line PO, which is a radius of the circle. We know that a tangent of a circle will have a radius which also meets that point and that those lines must meet at 90 degrees...



Two lines which are perpendicular to each other will have the negative inverse gradient of each other. Seeing as we are given the point (-1,4), we know that particular radius which meets the tangent will have a gradient of 4/-1, or just minus 4, as seen by the blue line above. Therefore, to get the gradient of the red line we want, swap 4/-1 round to become -1/4, but we need the negative of that negative, so the gradient of the red line is positive 1/4.