Preliminary Maths Question (1 Viewer)

[Crosswinds]

Okay, I have a stupid question, sorry.

How do you find the equation of a line passing through the point of intersection of two other lines and with a given gradient? I'm in year 10 and i don't think we've done this.

I know how to find a line passing through the intersection and another point but i can't do this.

Thankyou!

 

[lyounamu]

Okay, I have a stupid question, sorry.

How do you find the equation of a line passing through the point of intersection of two other lines and with a given gradient? I'm in year 10 and i don't think we've done this.

I know how to find a line passing through the intersection and another point but i can't do this.

Thankyou!

Let the point of intersection be (x1,y1) with the gradient being m.

Therefore, the equation of the line that passes through the point of intersection with the gradient m is given by:
y - y1 = m(x-x1)

If you want to find the point of intersection, work out the questions simultaneously to find the x-coordinates and y-coordinates. Gradient = rise/run or y-y1/x-x1.

 

[Crosswinds]

Okay thanks. I was hoping there was a quicker way, like there is with the other question i mentioned.

That would just be my optimistic laziness then.

 

[Aplus]

Okay, I have a stupid question, sorry.

How do you find the equation of a line passing through the point of intersection of two other lines and with a given gradient? I'm in year 10 and i don't think we've done this.

I know how to find a line passing through the intersection and another point but i can't do this.

Thankyou!

Just find the point of intersection of those two lines and use the point gradient formula.

 

[Aerath]

Okay thanks. I was hoping there was a quicker way, like there is with the other question i mentioned.

That would just be my optimistic laziness then.

How much quicker were you hoping for.

 

[Crosswinds]

Actually, I remember why I wanted a quicker way - it wasn't just my laziness. The question says don't find the coordinates of the point of intersection. Actually I'm just going to write the question (as i should have done in the first place, sorry).


"Let M be the point of intersection of the lines l1: 3x - 4y - 5 = 0 and l2: 4x + y + 7 = 0. Write down the equation of the general line through M. Hence, without actually finding the coordinates of M, find the equation of the line through M with gradient -3."

...And this is the answer:

"The general line through M has equation (3x - 4y - 5) + k(4x + y + 7) = 0. k = -15. 57x + 19y + 110 = 0."


I understand the equation of the general line, but i don't understand why k = -15 or therefore why the equation of the line is 57x + 19y + 110 = 0.


I'm soorrry!

 

[lyounamu]

Actually, I remember why I wanted a quicker way - it wasn't just my laziness. The question says don't find the coordinates of the point of intersection. Actually I'm just going to write the question (as i should have done in the first place, sorry).


"Let M be the point of intersection of the lines l1: 3x - 4y - 5 = 0 and l2: 4x + y + 7 = 0. Write down the equation of the general line through M. Hence, without actually finding the coordinates of M, find the equation of the line through M with gradient -3."

...And this is the answer:

"The general line through M has equation (3x - 4y - 5) + k(4x + y + 7) = 0. k = -15. 57x + 19y + 110 = 0."


I understand the equation of the general line, but i don't understand why k = -15 or therefore why the equation of the line is 57x + 19y + 110 = 0.


I'm soorrry!

That's a different case because you have general equations for each line.

In that case, you can say that

(equation of the first line) + k(equation of the second line) = 0

because 0 + 0 . k = 0 isn't it?

So M has equation (3x - 4y - 5) + k(4x + y + 7) = 0
Expand: 3x - 4y - 5 + 4kx + ky + 7k = 0
(k-4)y + (3+4k)x + 7k-5 = 0 (collect the like terms)
Make this general equation into the gradient form like this:
y = -((3+4k)/(k-4))x - (7k-5)/(k-4)
Since M has gradient of -3, -3 = -((3+4k)/(k-4))
Then -3 - 4k = -3k + 12
Therefore, k = -15
Then substitute that K value in to find the whole equation.

 

[Crosswinds]

Oohhh I get it I get it! This is one of the advantages of being terrible at maths - you have licence to be excited whenever something works.

Thankyou very much!

 

[Aerath]

Yet you're learning stuff ahead. You can't be -that- bad.

 

[lyounamu]

To OP: I would not really recommend anyone learning stuff ahead by themselves unless they actually get help along the way. I struggled so much last year. I had to go through 3 years of maths virtually with no help and I don't think anyone would like to spend 1-2 hours a day doing maths just to catch up.

It's always good to flow with the current program or through acceleration program where you can actually get help.

 

[Aplus]

If you want a faster way, don't use K method lol.

 

[lyounamu]

If you want a faster way, don't use K method lol.

You are being sarcastic, right?

K method is faster in that situation.

 

[Aplus]

You are being sarcastic, right?

K method is faster in that situation.

But gradient is given. And simultaneous equation to find point of intersection takes like 30 secs.

 

[lyounamu]

But gradient is given. And simultaneous equation to find point of intersection takes like 30 secs.

But you cannot use the simultaneous equation (according to the question). And usually, the gradient is not given for the questions like that when you are given 2 general equations.

 

[Aplus]

But you cannot use the simultaneous equation (according to the question). And usually, the gradient is not given for the questions like that when you are given 2 general equations.

I never noticed where it said that you can't use that method for that question lol. But I wasn't saying this method was the fastest. Just in this case it would probably be.

 

[lyounamu]

I never noticed where it said that you can't use that method for that question lol. But I wasn't saying this method was the fastest. Just in this case it would probably be.

Yeah.

 

[Doctor Jolly]

To OP: I would not really recommend anyone learning stuff ahead by themselves unless they actually get help along the way. I struggled so much last year. I had to go through 3 years of maths virtually with no help and I don't think anyone would like to spend 1-2 hours a day doing maths just to catch up.

It's always good to flow with the current program or through acceleration program where you can actually get help.

I've been going ahead of my maths class since I started High School. I'm actually about 3-4 Chapters ahead of my class right now. But I only do this for 2U Maths 3U is too scary to do by yourself.

 

[Aplus]

If you want to learn ahead you're better off doing it with a tutor or with an experienced person. It's a big risk if you're going to end up learning the wrong concepts...

 

[Iruka]

You know this k-method thingy, I was reading a book the other day (A course in pure mathematics by Hardy) which had an analogous thing for circles, ie., you have two intersecting circles,

(x-a)^2 + (y-b)^2 = r^2

and

(x-c)^2 + (y-d)^2 = R^2,

say (obviously there are some conditions on a, b, c, d, r, and R or the circles don't intersect), then the equation of any other circle that passes through the points of intersection of the first two circles is given by


(x-a)^2 + (y-b)^2 - r^2 + k[(x-c)^2 + (y-d)^2 - R^2] = 0,

and if you set k = -1, then you get the equation of the chord of contact of the circles (which, although it is a straight line, I guess you can consider it a circle of infinite radius).

Neat, huh.

 

[Doctor Jolly]

If you want to learn ahead you're better off doing it with a tutor or with an experienced person. It's a big risk if you're going to end up learning the wrong concepts...

Which is why I leave 3U stuff to my tutor. But the bad thing about going ahead is that you sometimes forget the things you learn because you're so ahead ... well that happens to me and my fish memory