General form of an equation with undefined slope

[Orson]

Homework Statement

The equation of the line that goes through the points

and

can be written in general form

. What are the values for A, B, and C?

Homework Equations

Ax+By+C=0

The Attempt at a Solution

[/B]
I think the answer is x=1. But how do i express that in general form?

 

[Mark44]

Homework Statement

The equation of the line that goes through the points View attachment 210311 and View attachment 210312 can be written in general form View attachment 210313 . What are the values for A, B, and C?

Homework Equations

Ax+By+C=0

The Attempt at a Solution

[/B]
I think the answer is x=1. But how do i express that in general form?

As x - 1 = 0.
In this form, A = 1. Can you figure out what B and C are?

 

[Orson]

The software says A=1 is wrong.

 

[Orson]

I just found out how to do it. Pretend you're trying to figure out the slope. you get 4/0 . 4 is the value for A. 0 for B. solve for c.

A= 4
B= 0
C= -4

 

[SammyS]

The software says A=1 is wrong.

Does the software mark A, B, and C individually or does it mark them together as a group of numbers ?

 

[Orson]

Does the software mark A, B, and C individually or does it mark them together as a group of numbers ?

Individually.
A= blank. B=blank. C=blank

 

[MarkFL]

I recommend reviewing post #2 to get the line in the given general form...

 

[SammyS]

I just found out how to do it. Pretend you're trying to figure out the slope. you get 4/0 . 4 is the value for A. 0 for B. solve for c.

A= 4
B= 0
C= -4

I take that the software says those answers are correct.

As it turns out there are infinitely many correct solutions. (You might say, infinitely many.)

The only one of the three which must have a particular value is B, which is zero.

The only requirements on A and C is that A = −C , so that A + C = 0 .

 

[Orson]

I take that the software says those answers are correct.

As it turns out there are infinitely many correct solutions. (You might say, infinitely many.)

The only one of the three which must have a particular value is B, which is zero.

The only requirements on A and C is that A = −C , so that A + C = 0 .

I am pretty sure at one point I had 1 and -1 for a and c and software marked it wrong. Grrrrrr.

 

[Orson]

I plugged 4 in for A and 0 for B because I read somewhere you act like you're going to figure the slope out (even though you can't) in order to get A and B.

 

[Nidum]

Rewrite the equation in slope and intercept form : y = mx + d

Sketch the graph of the line .

Is there something about the coordinate values given for those two points on the line that might cause you to think about this problem in a particular way ?

 

[SammyS]

I plugged 4 in for A and 0 for B because I read somewhere you act like you're going to figure the slope out (even though you can't) in order to get A and B.

Using that same idea, you can just as well get A = −4, B = 0, and C = 4 , by using the coordinates in the reverse order.

 

[Orson]

Using that same idea, you can just as well get A = −4, B = 0, and C = 4 , by using the coordinates in the reverse order.

Yes. I see how infinitely many solutions are possible for a and c. I put the equation in

Using that same idea, you can just as well get A = −4, B = 0, and C = 4 , by using the coordinates in the reverse order.

Yes. I see you can put any number and its complement to make b 0. I also solved for x and got x=1 which makes sense. How would i get the equation for a vertical line at 2?

 

[Orson]

Rewrite the equation in slope and intercept form : y = mx + d

Sketch the graph of the line .

Is there something about the coordinate values given for those two points on the line that might cause you to think about this problem in a particular way ?

They form a straight vertical line?

 

[Orson]

Rewrite the equation in slope and intercept form : y = mx + d

Sketch the graph of the line .

Is there something about the coordinate values given for those two points on the line that might cause you to think about this problem in a particular way ?

Also, I converted it to slope intercept form. I get 0=-4x+4. Is that what you're getting at?

 

[Orson]

Also, I converted it to slope intercept form. I get 0=-4x+4. Is that what you're getting at?

Or simply y=-4x+4?

 

[SammyS]

Yes. I see how infinitely many solutions are possible for a and c. I put the equation in

Yes. I see you can put any number and its complement to make b 0. I also solved for x and got x=1 which makes sense. How would i get the equation for a vertical line at 2?

So you want an equation of the form $\ Ax+By+C = 0\$ that with suitable values for A, B, C is equivalent to the equation $\ x=2\ .$

$\ x=2\$
is equivalent to
$\ x-2=0\ .$
Multiplying by A and adding $\ 0y\$ gives
$\ Ax+0y-2A=0\ .\$ (Solve it for x to convince yourself.)

So B = 0, and you can choose any non-zero number for A (you need x in the equation). Then C is −2 times the value chosen for A.

Otherwise, to do it by your method:
Pick any two distinct points, both having 2 for the x coordinate and proceed.

 

[epenguin]

Now you have got what Nidum #10 hinted. You could have said at the start: to one x value corresponds two y values; for a straight line that can only be a vertical line.
Sure you copied out the question right?

 

[Orson]

I am sure I copied the question right. Thanks.

 

[epenguin]

So in conclusion what is the equation of the line in simplest form?

 

[Orson]

So in conclusion what is the equation of the line in simplest form?

simplest form, I would say x=1. I am heeding what it says in your sig.

 

[epenguin]

Yes I think that is correct.
You may not yet have the reflexes that would have told you to see this straightaway. But in that case the suggestion of Nidum to sketch or plot the points is one that will come in useful in future not only for this problem, a generally useful hint.
Thank you for noting my sig , not everybody does.

 

[Mark44]

Also, I converted it to slope intercept form. I get 0=-4x+4. Is that what you're getting at?

Or simply y=-4x+4?

No, the latter equation is incorrect. This would be a line with slope of -4, passing through the point (0, 4). This is not a vertical line.

 

[Mark44]

The software says A=1 is wrong.

The the software is brain-dead, at least in this case.

 

[Orson]

No, the latter equation is incorrect. This would be a line with slope of -4, passing through the point (0, 4). This is not a vertical line.

I should have seen that.