Data Sufficiency

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]

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

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 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 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?

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?

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

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.