Maths help (2 Viewers)

[Shazer2]

How do I prove that 3 lines are concurrent? x-5y-17=0, 3x-2y-12=0 and 5x+y-7=0.

 

[Carrotsticks]

Solve two of them simultaneously and find a point.

Show that point satisfies the 3rd line.

 

[Shazer2]

If I have 4 equations, do I simultaneously solve 2 and then sub the meeting point into the other 2?

 

[Carrotsticks]

Yep!

 

[Shazer2]

Alright CarrotSticks, I got that sorted. I'm powering through this maths! Might have a few upcoming questions for this last exercise

 

[Carrotsticks]

Ask away!

 

[Shazer2]

If I need to find (based on 3 equations) that the vertices form a right-angle, do I need to solve them simultaneously to get 3 points, and find it the gradient from one to the other is perpendicular?

 

[Drongoski]

Not really. You need only demonstrate 2 lines are mutually perpendicular and that the 3rd line is not parallel to either of the 2 perp lines (and are not concurrent).

 

[Shazer2]

I need to find the gradients of 2 lines, and then multiply to get -1 to prove it's a right-angle triangle, correct?



So, that's a quick diagram, now the gradient from AB is -3 and 2/3, BC is -4/7 and AC is 1 and 3/4. If you multiply AC and BC, you result in -1. Is this right?

 

[qwerty44]

I need to find the gradients of 2 lines, and then multiply to get -1 to prove it's a right-angle triangle, correct?

Yep that proves those two are perpendicular to each other. Then show the third line crosses both of them to form vertices of a right-triangle.

 

[twinklegal19]

I need to find the gradients of 2 lines, and then multiply to get -1 to prove it's a right-angle triangle, correct?

View attachment 25655

So, that's a quick diagram, now the gradient from AB is -3 and 2/3, BC is -4/7 and AC is 1 and 3/4. If you multiply AC and BC, you result in -1. Is this right?

if you multiply the gradients of AC and BC, then yeah it'll be -1

it looks right

 

[RealiseNothing]

if you multiply the gradients of AC and BC, then yeah it'll be 1

it looks right

m8

 

[twinklegal19]

m8

aww shit

(y do u do dis)

 

[RealiseNothing]

On second thought, probably should stick to 2U methods.

 

[twinklegal19]

really? I never knew that

edit: just looked it up, is it called Thales' theorem?
edit2: oops nvm
edit3: oh I get it

 

[RealiseNothing]

really? I never knew that

edit: just looked it up, is it called Thales' theorem?
edit2: oops nvm

lol

 

[twinklegal19]

lol

do you mean this?

 

[RealiseNothing]

do you mean this?

If you prove MC = MB = MA then it is a right angled triangle.

 

[Shazer2]

Well, I got my results back for the topic test on most of the stuff I required help with and I got 26/48. It's not exactly what I was hoping to achieve, but it's a pass so that's ok.

One guy got 100%