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X intercept

by rahulvsd » Sun Mar 04, 2012 4:04 am
In the xy-plane if line m has negative slope and passes through the point (t,-4) is the x intercept of line m positive?

1. t = 0
2. The y interept if line m is negative.

[spoiler]OA: D[/spoiler]
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by pemdas » Sun Mar 04, 2012 12:20 pm
rahulvsd wrote:In the xy-plane if line m has negative slope and passes through the point (t,-4) is the x intercept of line m positive?

1. t = 0
2. The y interept if line m is negative.

[spoiler]OA: D[/spoiler]
i solved this question graphically for answer d
the line has negative slope and st(1) says t=0 which means that y-intercept is -4. Hence the line must have negative x-intercept. The only way when with (0,-4) the x-intercept is not negative is when we have undefined slope i.e. slope=delta y/0 or positive slope.
st(2) y-intercept is negative implies x-intercept must be negative too for exactly the same reasoning explained in st(1) plus it would be possible for x-intercept to be not defined when the slope=0 (the line is horizontal, y-intercept=!0), but we are given that the slope is negative in the question.
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by krusta80 » Sun Mar 04, 2012 2:18 pm
rahulvsd wrote:In the xy-plane if line m has negative slope and passes through the point (t,-4) is the x intercept of line m positive?

1. t = 0
2. The y interept if line m is negative.

[spoiler]OA: D[/spoiler]
Let's translate this word problem into algebra by using the formula for line M

y = mx + b, where this m is the slope and b is the y-intercept
m < 0
-4 = mt + b -> t = (-b-4)/m = -b/m - 4/m

To get the x-intercept, set y = 0
x = -b/m = t + 4/m

Is -b/m > 0?

Part (1)
t = 0

Therefore, x-intercept = 4/m. Since m is negative, we know that the x-intercept is NOT greater than 0. SUFFICIENT

Part (2)
b < 0

x-int = -b/m. Again, the x-intercept will always be negative, since both b and m are negative, and we are negating their quotient. SUFFICIENT

D
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by pemdas » Mon Mar 05, 2012 4:01 am
hi krusta, i spotted some minor issues in your solution and decided to feedback.
Please follow my analysis carefully before standing with your approach affirmed :)

The bold green part, here you set y=0 and find x-intercept - agree
y=mx+b and y=0 <> 0=mx+b, x=-b/m

The bold blue part, here you supply -4 for y from the given coordinate (t,-4) - agree, BUT this will not be equivalent to the expression for x-intercept above (bold green part) as y=!0 and y=-4. Next, you consider t= -b/m - 4/m and keep this equivalent (mistakenly at this time) to x=-b/m. For the latter (x=-b/m) y=0 and not y=-4. Do you see this?

Instead I would proceed from this part right to the statement consideration
Let's translate this word problem into algebra by using the formula for line M

y = mx + b, where this m is the slope and b is the y-intercept
m < 0
-4 = mt + b -> t = (-b-4)/m


st(1) t=0 and the only way when (-b-4)/m=0 with the negative slope m (not 0) is when b=-4. Put this on graph and see that it's only possible with negative x-intercept. For the more detailed reasoning see the post with my graphical solution.

for the st(2) again I would put through my reasoning above for the graphical solution.

p.s. Based on my own experience I could say that an algebraic solution of slope and line equation problem(s) is somewhat tedious. Therefore I always put everything on graph and analyze.
krusta80 wrote:
rahulvsd wrote:In the xy-plane if line m has negative slope and passes through the point (t,-4) is the x intercept of line m positive?

1. t = 0
2. The y interept if line m is negative.

[spoiler]OA: D[/spoiler]
Let's translate this word problem into algebra by using the formula for line M

y = mx + b, where this m is the slope and b is the y-intercept
m < 0
-4 = mt + b -> t = (-b-4)/m = -b/m - 4/m


To get the x-intercept, set y = 0
x = -b/m = t + 4/m

Is -b/m > 0?

Part (1)
t = 0

Therefore, x-intercept = 4/m. Since m is negative, we know that the x-intercept is NOT greater than 0. SUFFICIENT

Part (2)
b < 0

x-int = -b/m. Again, the x-intercept will always be negative, since both b and m are negative, and we are negating their quotient. SUFFICIENT

D
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by krusta80 » Mon Mar 05, 2012 4:17 am
pemdas wrote:hi krusta, i spotted some minor issues in your solution and decided to feedback.
Please follow my analysis carefully before standing with your approach affirmed :)

The bold green part, here you set y=0 and find x-intercept - agree
y=mx+b and y=0 <> 0=mx+b, x=-b/m

The bold blue part, here you supply -4 for y from the given coordinate (t,-4) - agree, BUT this will not be equivalent to the expression for x-intercept above (bold green part) as y=!0 and y=-4. Next, you consider t= -b/m - 4/m and keep this equivalent (mistakenly at this time) to x=-b/m. For the latter (x=-b/m) y=0 and not y=-4. Do you see this?

Instead I would proceed from this part right to the statement consideration
Let's translate this word problem into algebra by using the formula for line M

y = mx + b, where this m is the slope and b is the y-intercept
m < 0
-4 = mt + b -> t = (-b-4)/m


st(1) t=0 and the only way when (-b-4)/m=0 with the negative slope m (not 0) is when b=-4. Put this on graph and see that it's only possible with negative x-intercept. For the more detailed reasoning see the post with my graphical solution.

for the st(2) again I would put through my reasoning above for the graphical solution.

p.s. Based on my own experience I could say that an algebraic solution of slope and line equation problem(s) is somewhat tedious. Therefore I always put everything on graph and analyze.
krusta80 wrote:
rahulvsd wrote:In the xy-plane if line m has negative slope and passes through the point (t,-4) is the x intercept of line m positive?

1. t = 0
2. The y interept if line m is negative.

[spoiler]OA: D[/spoiler]
Let's translate this word problem into algebra by using the formula for line M

y = mx + b, where this m is the slope and b is the y-intercept
m < 0
-4 = mt + b -> t = (-b-4)/m = -b/m - 4/m


To get the x-intercept, set y = 0
x = -b/m = t + 4/m

Is -b/m > 0?

Part (1)
t = 0

Therefore, x-intercept = 4/m. Since m is negative, we know that the x-intercept is NOT greater than 0. SUFFICIENT

Part (2)
b < 0

x-int = -b/m. Again, the x-intercept will always be negative, since both b and m are negative, and we are negating their quotient. SUFFICIENT

D
Pemdas,

I am able to combine the blue and green because the "blue" equation has neither an x nor a y in it. In other words, the slope, m, and y-intercept, b, (as well t) remain constant for all points (x,y) on the line, which means I am well within the laws of geometry to perform this substitution.

Hopefully this clears things up for you.
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by pemdas » Mon Mar 05, 2012 5:28 am
you are equating one point in space (x-y) to another point in space (x-y). Nobody says you cannot keep the same line equation with slope for the different coordinates. But to equate two coordinates as one is not allowed. Care to reconsider?
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by krusta80 » Mon Mar 05, 2012 7:09 am
pemdas wrote:you are equating one point in space (x-y) to another point in space (x-y). Nobody says you cannot keep the same line equation with slope for the different coordinates. But to equate two coordinates as one is not allowed. Care to reconsider?
Perhaps the source of confusion is the following line from my original reply:

To get the x-intercept, set y = 0
x = -b/m = t + 4/m

The "x" above is stricly referring to the x intercept.
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by pemdas » Mon Mar 05, 2012 8:57 am
yes exactly, not confusion- minor issue to which I referred initially x= -b/m =! t+ 4/m because you obtained x=-b/m when y=0 and could obtain x=t+4/m when y=-4. Do you see this?
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by krusta80 » Mon Mar 05, 2012 9:04 am
pemdas wrote:yes exactly, not confusion- minor issue to which I referred initially x= -b/m =! t+ 4/m because you obtained x=-b/m when y=0 and could obtain x=t+4/m when y=-4. Do you see this?
Ah, I finally found your mistake...hopefully we can put this to rest now:

You said that I obtained x = t + 4/m, but I did not...

I obtained t = (-b-4)/m = -b/m - 4/m

And again, the "x" I was referring to was the x-intercept, which is equal to -b/m. So, plugging in the x-intercept for -b/m into the above equation, I got:

t = (x-intercept) - 4/m
(x-intercept) = t + 4/m

Again, I should have probably dedicated a different letter (maybe c?) for x intercept. But my logic and method of solving are perfectly correct.
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by pemdas » Mon Mar 05, 2012 9:33 am
this is last post i enter, as this debate doesn't prove constructive

if something comes up, I ask experts to comment on this thread
krusta80 wrote:
rahulvsd wrote:In the xy-plane if line m has negative slope and passes through the point (t,-4) is the x intercept of line m positive?

1. t = 0
2. The y interept if line m is negative.

[spoiler]OA: D[/spoiler]
Let's translate this word problem into algebra by using the formula for line M

y = mx + b, where this m is the slope and b is the y-intercept
m < 0
-4 = mt + b -> t = (-b-4)/m = -b/m - 4/m
 pemdas query: if y=-4 and y=mt+b is line equation, then -4=mt+b holds true and t=-4/m -b/b according to you and i agree
To get the x-intercept, set y = 0
x = -b/m = t + 4/m pemdas query: if y=0 and y=mt+b is line equation, then 0=mt+b holds true and t=-b/m according to me BUT you put x= -b/m = t + 4/m which is the same as x= -b/m + 4/m. The value of 't' obtained when y=0 and y=-4 are different, because 't' is 'x' here and two distinctive points cannot have the same x-coordinate and distinctive y-coordinates when the slope is defined, i.e. the line is not straight vertical line

Is -b/m > 0?

Part (1)
t = 0

Therefore, x-intercept = 4/m. Since m is negative, we know that the x-intercept is NOT greater than 0. SUFFICIENT

Part (2)
b < 0

x-int = -b/m. Again, the x-intercept will always be negative, since both b and m are negative, and we are negating their quotient. SUFFICIENT

D
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by Mike@Magoosh » Mon Mar 05, 2012 12:26 pm
To this intelligent discussion, I would like to add my 2 cents. :)

First of all, I will say, it's always an honor to join a discussion in which such high level mathematics is being discussed. My respect for all who have posted in this thread.

Having said that, I will say that I want to defend krusta80's correct approach from my friend pemdas' objections.

So, if I understand krusta80's approach, we started with the general formula for a line . . .

y = mx + b

with the understanding that m < 0. Notice, the constants "m" and "b" are true for all points on the line.

We plugged in the point (t, -4), and got

-4 = mt + b ---> t = -b/m - 4/m

Now, this is subtle. We created that equation by plugging in the the point (t, -4), but once created, this is an algebraic relationship among the three variables that is independent of any particular point. In this equation, "m" is the slope at every point, "b" is the "y-intercept", true for the line as a whole, and "t" is a variable about which we don't know much. We know "t" is the x-coordinate when y = -4, and whether t plays some other role remains to be seen. One cannot afford to be an algebraic fundamentalist. Just because t is the y-value of one point doesn't mean it can't be anything else. Just because a variable or relationship arises in one context, that does not limit the scope of the algebra. The very point of algebra is to translate into a symbolic form that exposes the relationships among the variables that are independent of context.

In my reading, krusta80 then went to the x intercept by setting the y equal to zero

x-int = -b/m

Then, very cleverly, krusta80 noticed that the green equation above could be reconjiggered, to make -b/m = t + 4/m

Thus,

x-int = t + 4/m

Again, it matters not at all that the green equation was established by plugging in another point. We established a relationship among the variables, and such a relationship is independent of any single point. We then applied that relationship to the expression for the y-intercept.

Admittedly, this is a somewhat strange equation, this indigo one --- it relates the x-intercept to the slope and to t (the y-coordinate when x = -4). Strange, but perfectly valid, and completely universal. You will find that, for any oblique line in the x-y plane, if you plug in the slope m and t (the y-coordinate when x = -4), this equation will always give you the x-intercept of the line. It works for all infinity of the possible oblique lines in the plane. An idiosyncratic but nonetheless extraordinary result.

The rest of krusta80's analysis is also 100% correct, and I believe that part is uncontroversial, so I will not comment on it.

Please let me know if anyone reading this has any questions on what I've said here.

Mike :)
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by pemdas » Mon Mar 05, 2012 1:02 pm
ok, clear now x-intercept=t+4/m not x=t+4/m

thanks Mike and this is really strange equation here as krusta allows substitution of t=x only for x-intercept. I would say this is very special case valid for x-intercept only, correct me if i'm wrong.
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by krusta80 » Mon Mar 05, 2012 1:05 pm
pemdas wrote:ok, clear now x-intercept=t+4/m not x=t+4/m

thanks Mike and this is really strange equation here as krusta allows substitution of t=x only for x-intercept. I would say this is very special case valid for x-intercept only, correct me if i'm wrong.
My mistake was not being more careful differentiating x from x-intercept in my formula. What I did would work for any formula without variables in it. Note that like m for slope and b for y-intercept, x-intercept never changes for a line and therefore is not a variable. It just looks like one. :)
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