# finding lines

• Mar 23rd 2009, 11:31 AM
Keep
finding lines
Here is the question. Find two lines L1, L2 that intersect at point (a,b). So how do yo go about it? Any help is appreciated, thanks.
• Mar 23rd 2009, 11:40 AM
earboth
Quote:

Originally Posted by Keep
Here is the question. Find two lines L1, L2 that intersect at point (a,b). So how do yo go about it? Any help is appreciated, thanks.

Let \$\displaystyle m_1\$ denote the slope of \$\displaystyle L_1\$, \$\displaystyle m_2\$ the slope of \$\displaystyle L_2\$.

Then the equations of the lines are

\$\displaystyle L_1: y-b=m_1(x-a)\$ and

\$\displaystyle L_2: y-b=m_2(x-a)\$
• Mar 24th 2009, 10:54 AM
Keep
Quote:

Originally Posted by earboth
Let \$\displaystyle m_1\$ denote the slope of \$\displaystyle L_1\$, \$\displaystyle m_2\$ the slope of \$\displaystyle L_2\$.

Then the equations of the lines are

\$\displaystyle L_1: y-b=m_1(x-a)\$ and

\$\displaystyle L_2: y-b=m_2(x-a)\$

But you need to find out at least the slope of the line to get it, don't you? And since you don't have the y intercept of the line, that would also be difficult to get, isn't it? So you have two lines intersecting at point (6,18) and you don't have the gradient (slope) or the y intercept. You are just given the point of intersection of x and y where x = 6 and y = 18 as I said above.
• Mar 24th 2009, 11:47 AM
Keep
Can anyone help? I just have ONE point, i.e. the point of intersection of the two lines ONLY (no slope or y intercept given) and nothing else. SO how do I get their equations?
• Mar 24th 2009, 12:03 PM
Plato
\$\displaystyle L_1:~y-b=2(x-a)\$
\$\displaystyle L_2:~y-b=3(x-a)\$

There. You now have two lines.
• Mar 24th 2009, 12:13 PM
Keep
Quote:

Originally Posted by Plato
\$\displaystyle L_1:~y-b=2(x-a)\$
\$\displaystyle L_2:~y-b=3(x-a)\$

There. You now have two lines.

Thanks but I just want to understand more. 2 and 3 are the slopes for L1 and L2 respectively? If so, can you please tell me how you got because I want to know. So the equation will be y =2x-2a+b for L1 and y=3x-3a+b for L2? And what are a and b? Lastly, s it it not possible for us to get the equation in the form y=mx+b? Thank you and sorry for asking too many questions.
• Mar 24th 2009, 12:19 PM
Plato
Quote:

Originally Posted by Keep
Thanks but I just want to understand more. 2 and 3 are the slopes for L1 and L2 respectively? If so, can you please tell me how you got because I want to know.

I just made them up.
You can use any two numbers.
• Mar 24th 2009, 12:24 PM
Keep
Quote:

Originally Posted by Plato
I just made them up.
You can use any two numbers.

So 2 and 3 are the slopes of the lines that you get by choosing any two numbers, right? OK, I understand that, thanks. But what is a and b? b is the y intercept? If b is y intercept, then what is a?
• Mar 24th 2009, 12:30 PM
Plato
Quote:

Originally Posted by Keep
So 2 and 3 are the slopes of the lines that you get by choosing any two numbers, right? OK, I understand that, thanks. But what is a and b? b is the y intercept? If b is y intercept, then what is a?

If \$\displaystyle y-b=m(x-a)\$ then \$\displaystyle y=mx+(\color{blue}b-ma)\$.
So \$\displaystyle \color{blue}b-ma\$ is the y-intercept.
• Mar 24th 2009, 12:37 PM
Keep
Quote:

Originally Posted by Plato
If \$\displaystyle y-b=m(x-a)\$ then \$\displaystyle y=mx+(\color{blue}b-ma)\$.
So \$\displaystyle \color{blue}b-ma\$ is the y-intercept.

Ok but the equatio says that y intrecept is b minus the product of the slope and a. My question is, what do these two variables (a and b) stand for. I mean we know what x, y and m are but what are a and b? Lastly is it not possible for me to give the lines in the usual form of y=mx+b without b being equal to (b-ma)?
• Mar 24th 2009, 12:44 PM
Keep
Quote:

Originally Posted by Keep
Ok but the equatio says that y intrecept is b minus the product of the slope and a. My question is, what do these two variables (a and b) stand for. I mean we know what x, y and m are but what are a and b? Lastly is it not possible for me to give the lines in the usual form of y=mx+b without b being equal to (b-ma)?

Or maybe a and b are the points of intersection which I gave as 6 and 18?
• Mar 24th 2009, 12:46 PM
Plato
Quote:

Originally Posted by Keep
My question is, what do these two variables (a and b) stand for. ?

I have no idea what they stand for other that the coordinates of a point in the plane.
They simply were part of the given in the problem.
\$\displaystyle y=mx+(b-ma)\$ is in the form \$\displaystyle y=mx+B\$, where \$\displaystyle B=b-ma\$.
• Mar 24th 2009, 12:54 PM
Keep
Quote:

Originally Posted by Plato
I have no idea what they stand for other that the coordinates of a point in the plane.
They simply were part of the given in the problem.
\$\displaystyle y=mx+(b-ma)\$ is in the form \$\displaystyle y=mx+B\$, where \$\displaystyle B=b+mx\$.

Oh OK I get it now. I see. I know the first time I gave the coordinates as (a,b) but later on I put them as (6,18). So let us say we have the intersection point as (6,18). Using the equation which you gave which was:
\$\displaystyle L_1:~y-b=2(x-a)\$
\$\displaystyle L_2:~y-b=3(x-a)\$ we get:

L1: y=2x+6
L2: y=3x

Am I right?
• Mar 24th 2009, 01:39 PM
HallsofIvy
Do you understand that there are an infinite number of lines passing through the given point? You only asked for two of them. Any line passing through point \$\displaystyle (x_0, y_0)\$ can be written in the form \$\displaystyle y= m(x- x_0)+ y_0\$. Pick any two numbers for m and you get two different lines through \$\displaystyle (x_0,y_0)\$.
• Mar 24th 2009, 01:46 PM
Keep
Quote:

Originally Posted by HallsofIvy
Do you understand that there are an infinite number of lines passing through the given point? You only asked for two of them. Any line passing through point \$\displaystyle (x_0, y_0)\$ can be written in the form \$\displaystyle y= m(x- x_0)+ y_0\$. Pick any two numbers for m and you get two different lines through \$\displaystyle (x_0,y_0)\$.

Ok, thanks. So if I have a point of intersection of the two lines (L1 and L2) as (6,18) and I pick m=2 for L1 and m=3 for L2, the equation will be y=2(x-6)+18 for L1 and y=3(x-6)+18 for L2, and I will get y=2x+6 for L1 and y=3x for L2.