# Finding point-normal equations

## Homework Statement

Say that I have a problem along the lines of: Find the point-normal equation for the line L.

L=u1+k*u2, where k is ℝ.

And then I'm given u1 and u2, which both are 2×1 vectors.

Then L will also be a 2×1 vector (which consists of a11 and a21), which I guess is ok (since it gives the two coordinats in R2).

However the point-normal equation is in the form of (for some constants a and b and some points x0 and y0 on the line L):

a(x-x0)+b(y-0)=0.

So here comes my question: In the vector L: Are a11 or a21 x?

-

See above.

LCKurtz
Homework Helper
Gold Member

## Homework Statement

Say that I have a problem along the lines of: Find the point-normal equation for the line L.

L=u1+k*u2, where k is ℝ.

And then I'm given u1 and u2, which both are 2×1 vectors.

Then L will also be a 2×1 vector (which consists of a11 and a21), which I guess is ok (since it gives the two coordinats in R2).

However the point-normal equation is in the form of (for some constants a and b and some points x0 and y0 on the line L):

a(x-x0)+b(y-0)=0.

So here comes my question: In the vector L: Are a11 or a21 x?

-

## The Attempt at a Solution

See above.

Let ##u_1=\langle a, b\rangle## and ##u_2=\langle d_1,d_2\rangle##. Then your line is ##\langle x,y\rangle = \langle a, b\rangle + k\langle d_1,d_2\rangle##. Eliminate the ##k## between the ##x## and ##y## equations and put it in your required form. Note that ##(a,b)## is a point on the line.

• 1 person
Let ##u_1=\langle a, b\rangle## and ##u_2=\langle d_1,d_2\rangle##. Then your line is ##\langle x,y\rangle = \langle a, b\rangle + k\langle d_1,d_2\rangle##. Eliminate the ##k## between the ##x## and ##y## equations and put it in your required form. Note that ##(a,b)## is a point on the line.
Two questions:

1. (Perhaps a dumb question) Is ##\langle x,y\rangle## another way of writing a 2×1 matrix where x=a11 and y=a21? (I'm really new to linear algebra...)

2. If I'm correct about question 1, then I still don't see how I can "translate" ##\langle x,y\rangle = \langle a, b\rangle + k\langle d_1,d_2\rangle## to the form of

a(x-x0)+b(y-y0)=0

(That was what my original question was all about.)

Thanks for the help LCKurtz!

LCKurtz
Homework Helper
Gold Member
Two questions:

1. (Perhaps a dumb question) Is ##\langle x,y\rangle## another way of writing a 2×1 matrix where x=a11 and y=a21? (I'm really new to linear algebra...)

It does't matter whether you use a 2x1 or 1x2 matrix in this problem. You have two entries either way.

2. If I'm correct about question 1, then I still don't see how I can "translate" ##\langle x,y\rangle = \langle a, b\rangle + k\langle d_1,d_2\rangle## to the form of

a(x-x0)+b(y-y0)=0

(That was what my original question was all about.)

Thanks for the help LCKurtz!

Do what I suggested. Write the equations for ##x## and ##y## from the given matrix equation and eliminate the ##k## parameter. Try it.

It does't matter whether you use a 2x1 or 1x2 matrix in this problem. You have two entries either way.

Do what I suggested. Write the equations for ##x## and ##y## from the given matrix equation and eliminate the ##k## parameter. Try it.
I'm not sure what you mean with "eliminate the ##k## parameter".

But I guess this is a start?

##x=a+kd_1##

and

##y=b+kd_2##

now let me put ##x## and ##y## into the "form" of the point-normal equation. Which I will call:

##p(x-x_0)+q(y-y_0)=0##, where ##p## and ##q## are constants.

We know that ##x_0=a## and ##y_0=b## are solutions because ##k## could be ##0##.

so

##p((a+kd_1)-a)+q((b+kd_2)-b)=0##

##p(kd_1)+q(kd_2)=0##

##k(pd_1+qd_2)=0##

##pd_1+qd_2=0##

where ##(p,d)## is the normal to ##(x,y)##. But how do you get ##(p,d)##?

Does the calculations look ok LCKurtz?

Last edited:
LCKurtz
Homework Helper
Gold Member
Not quite. From ##x=a+kd_1,~y=b+kd_2## you eliminate ##k##. Solve each equation for ##k## and set them equal:$$k = \frac {x-a}{d_1} = \frac {y-b}{d_2}$$Put that last equation in the form$$(?)(x-a) + (??)(y-b) = 0$$and you will have your form.

• 1 person
I'm a bit lost. :/ Any more tips?

Not quite. From ##x=a+kd_1,~y=b+kd_2## you eliminate ##k##. Solve each equation for ##k## and set them equal:$$k = \frac {x-a}{d_1} = \frac {y-b}{d_2}$$Put that last equation in the form$$(?)(x-a) + (??)(y-b) = 0$$and you will have your form.
I know that the ##(?)## ##(??)## are the normal to the line ##\langle x,y\rangle = \langle a, b\rangle + k\langle d_1,d_2\rangle##. However what I don't know is how you should obtain the normal for that line...

HallsofIvy
Homework Helper
You say you are taking linear algebra but the problems you are having are all much more basic algebra matters.

$$\frac{x-a}{d_1}= \frac{y- b}{d_2}$$

An obvious thing to do is to multiply both sides by $d_1d_2$ to get rid of the fractions:
$$d_2(x- a)= d_1(y- b)$$
$$d_2(x- a)- d_1(y- b)= 0$$

You say you are taking linear algebra but the problems you are having are all much more basic algebra matters.

$$\frac{x-a}{d_1}= \frac{y- b}{d_2}$$

An obvious thing to do is to multiply both sides by $d_1d_2$ to get rid of the fractions:
$$d_2(x- a)= d_1(y- b)$$
$$d_2(x- a)- d_1(y- b)= 0$$
I feel like such an idiot now, that was my initial thought but I dismissed it because it felt like it was too easy to be the right answer. 