Equation of A Line

harpazo

harpazo

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An equation of the line containing the two points (x_1, y_1) and (x_2, y_2) may be expressed as the determinant shown in the picture.

20190925_113238.jpg

Prove this result by expanding the determinant and comparing the result to the two-point form of the equation of the line.

1. Seeking the first-two steps and hints.

2. What exactly is the purpose of this question in a Sullivan college algebra textbook? Is this prove more for a linear algebra course?

3. Use the result here to answer the next thread titled Collinear Points.
 
harpazo said:
An equation of the line containing the two points (x_1, y_1) and (x_2, y_2) may be expressed as the determinant shown in the picture.

View attachment 13853

Prove this result by expanding the determinant and comparing the result to the two-point form of the equation of the line.

1. Seeking the first-two steps and hints.

2. What exactly is the purpose of this question in a Sullivan college algebra textbook? Is this prove more for a linear algebra course?

3. Use the result here to answer the next thread titled Collinear Points.
Click to expand...
1. Seeking the first-two steps and hints. ................... calculate the determinant and put it in the form:

(y - y1) / (y2 - y1) = (x - x1) / (x2 - x1)

This is the equation of a line passing through (x1,y1) and (x2,y2)
 
From the Two-Point form of a line, posted by Subhotosh, we can also "see" the Point-Slope form. Multiply each side by y2-y1.

(y - y1) / (y2 - y1) = (x - x1) / (x2 - x1)

(y2 - y1) * (y - y1) / (y2 - y1) = (y2 - y1) * (x - x1) / (x2 - x1)

y - y1 = (x - x1) * (y2 - y1) / (x2 - x1)

Do you remember the Slope Formula (the difference of y coordinates over the difference of x coordinates)?

m = (y2 - y1) / (x2 - x1)

Replacing the slope with symbol m (in the third equation above) yields:

y - y1 = m * (x - x1)

That result is also the equation of a line passing through (x1,y1) and (x2,y2). We call it the Point-Slope Formula. (We use it to write the equation of a line when we know both the slope and the coordinates of one point on the line. That's why it's called Point-Slope form.)

If you forget the Two-Point form, you can recover it from the Point-Slope form, by working the steps above in reverse.

?
 
Subhotosh Khan said:
1. Seeking the first-two steps and hints. ................... calculate the determinant and put it in the form:

(y - y1) / (y2 - y1) = (x - x1) / (x2 - x1)

This is the equation of a line passing through (x1,y1) and (x2,y2)
Click to expand...

Good start for me.
 
Otis said:
From the Two-Point form of a line, posted by Subhotosh, we can also "see" the Point-Slope form. Multiply each side by y2-y1.

(y - y1) / (y2 - y1) = (x - x1) / (x2 - x1)

(y2 - y1) * (y - y1) / (y2 - y1) = (y2 - y1) * (x - x1) / (x2 - x1)

y - y1 = (x - x1) * (y2 - y1) / (x2 - x1)

Do you remember the Slope Formula (the difference of y coordinates over the difference of x coordinates)?

m = (y2 - y1) / (x2 - x1)

Replacing the slope with symbol m (in the third equation above) yields:

y - y1 = m * (x - x1)

That result is also the equation of a line passing through (x1,y1) and (x2,y2). We call it the Point-Slope Formula. (We use it to write the equation of a line, when we know the coordinates of one point on the line and know also the line's slope. That's why it's called Point-Slope form.)

If you forget the Two-Point form, you can recover it from the Point-Slope form, by working the steps above in reverse.

?
Click to expand...

Nice reply. Thanks again.
 
Subhotosh Khan said:
1. Seeking the first-two steps and hints. ................... calculate the determinant and put it in the form:

(y - y1) / (y2 - y1) = (x - x1) / (x2 - x1)

This is the equation of a line passing through (x1,y1) and (x2,y2)
Click to expand...

I need additional steps to solve the problem.
 
harpazo said:
[MATH][/MATH]

See picture for the start of the problem.
Note: I cannot answer the Collinear Points problem without showing this prove.
View attachment 13899
Click to expand...
No! That should be:
x * [y1 - y2] - y * [x1-x2] + [x1 * y2 - x2 * y1] = 0

Fix that. Then rearrange to get:

[ x2* y - x1* y - x2*y1 + x1*y1] - [x * y2 - x * y1 - x1*y2 + x1*y1] = 0

Continue........
 
Subhotosh Khan said:
No! That should be:
x * [y1 - y2] - y * [x1-x2] + [x1 * y2 - x2 * y1] = 0

Fix that. Then rearrange to get:

[ x2* y - x1* y - x2*y1 + x1*y1] - [x * y2 - x * y1 - x1*y2 + x1*y1] = 0

Continue........
Click to expand...

Forget it! Moving on. I should not be trying to solve math problems that are far above my head.
 
You said you never give up. And that wouldn't be moving on.

This is not above your head. You're just making little mistakes, probably by writing too fast or not comparing each line you write to what you intend.

I suspect you just need more sleep or something. Be patient, you'll get there.
 
Dr.Peterson said:
You said you never give up. And that wouldn't be moving on.

This is not above your head. You're just making little mistakes, probably by writing too fast or not comparing each line you write to what you intend.

I suspect you just need more sleep or something. Be patient, you'll get there.
Click to expand...

More sleep is right. I work out all math work after my overnight work. I will blame all typos on lack of sleep.
 
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