arora007 wrote:Lines n and p lie on the xy plane. Is the slope of line n less than slope of line p?
(1) Lines n and p intersect at (5, 1)
(2) The y-intercept of line n is greater than the y intercept of p


in general i am not very strong at co-ordinate geometry.
If you somebody could please help me visualize the above problem?

what does slope less than or more than mean...maybe with a figure or quadrant wise.

OA C

I have added a figure which has lines 1 , 2, 3 , 4

can 1 & 3 make a pair too ?
or is it strictly 1,2 and 3,4 and which is n and which is p ??

please do explain what is negative and postive slopes as well...

(1) Coordinates of the point of intersection of two non parallel lines in a plane tell nothing about the slope(s) of either line. Insufficient

(2) Since, slope of a line is (-) y-intercept/x-intercept, the smaller the y-intercept the greater the slope of the line for a given positive x-intercept, read opposite for a given negative x-intercept; which is no where mentioned. Insufficient

When blended

Lines n and p intersecting at (5, 1), and the y-intercept of line n is greater than the y intercept of line p. This raises too many more questions to be answered before answering to the stem; like whether the y-intercept of line n is positive, a zero, or negative, and similar questions about the x-intercept(s). Which, if studied, could take us to a decision if you view the following post by arora007, who made me to rethink and edit my work.

[spoiler]C[/spoiler]

Hi well..this is the text from papgust's 300 question set... but after a point I did not understand a word....
btw I am not as interested in the question as much I am interested in the concept!

Statement 1 tells us where the lines intersect, but tells us nothing about either of the lines' slopes. If you draw a picture of two lines intersecting at (5,1), you can see that with the information given, we could label either one n. We could make the one with greater slope n, or the one with less slope n. So, we don't have enough information from statement 1.
Now let's consider statement 2, that says that the y intercept of n is greater than that of the y intercept of p. The slopes could be unequal (think about intersecting lines), or we could have parallel lines, in which case the slopes are equal. So, statement 2 is not sufficient on its own.

Now let's consider the statements together. The lines intersect at (5,1), and n has the higher y intercept. Let's look at 3 cases:
1) Both have y intercepts above y=1
Since n intersects higher, then we know n had further to descend, so its slope is steeper (but more negative) than p's. Thus, p has a greater slope.
2) n has intercept above y=1, p has intercept below
n would have a negative slope and p a positive, so p has a greater slope
3) Both have y intercepts below y=1
Both have positive slopes, but p has further to ascend. Thus, p has a greater slope.

Since combining the information tells us that p always has a greater slope, we have sufficient information with both statements and the answer is C.

i am also weak in geometry
but my little try

n=y=k1*x+b1
line p
p=y=k2*x+b2
here k1 and k2 slopes of the lines, and b1, b2 y-intercepts
we need to know does k1>k2
or does
(y-b1)/x>(y-b2)/x
(1) we are given the value of x and y (5,1) but know the value of b, insufficient
(2) we are given that b1>b2 but no info about x,y insufficient
both we are given the value of x,y and that b1>b2 so sufficient

clock60 wrote:i am also weak in geometry
but my little try

n=y=k1*x+b1
line p
p=y=k2*x+b2
here k1 and k2 slopes of the lines, and b1, b2 y-intercepts
we need to know does k1>k2
or does
(y-b1)/x>(y-b2)/x
(1) we are given the value of x and y (5,1) but know the value of b, insufficient
(2) we are given that b1>b2 but no info about x,y insufficient
both we are given the value of x,y and that b1>b2 so sufficient

awesome

clock60 wrote:i am also weak in geometry
but my little try

n=y=k1*x+b1
line p
p=y=k2*x+b2
here k1 and k2 slopes of the lines, and b1, b2 y-intercepts
we need to know does k1>k2
or does
(y-b1)/x>(y-b2)/x
(1) we are given the value of x and y (5,1) but know the value of b, insufficient
(2) we are given that b1>b2 but no info about x,y insufficient
both we are given the value of x,y and that b1>b2 so sufficient

well.. algebrically... this makes sense... but can you please help
I have added a figure which has lines 1 , 2, 3 , 4

can 1 & 3 make a pair too ?
or is it strictly 1,2 and 3,4 and which is n and which is p ??

please do explain what is negative and postive slopes as well...

arora007 wrote:well.. algebrically... this makes sense... but can you please help
I have added a figure which has lines 1 , 2, 3 , 4

can 1 & 3 make a pair too ?
or is it strictly 1,2 and 3,4 and which is n and which is p ??

please do explain what is negative and postive slopes as well...

1 and 3 definitely make a pair and so do 2 and 4. Here 1 and 2 have a negative slope as their angles with positive x-axis are each obtuse, and by definition, slope of a line that makes α angle with positive x-axis is given by tan α, which turns out to be negative when 90° < α < 180°, and positive when 0° < α < 90°. Revise "Add Sugar To Coffee" concept of Trigonometry for a better understanding.

In 1 and 2, 1 is n. In 3 and 4, 3 is n. In 1 and 3, 1 is n. In 2 and 4, 2 is n.

sanju09 wrote:

arora007 wrote:well.. algebrically... this makes sense... but can you please help
I have added a figure which has lines 1 , 2, 3 , 4

can 1 & 3 make a pair too ?
or is it strictly 1,2 and 3,4 and which is n and which is p ??

please do explain what is negative and postive slopes as well...

1 and 3 definitely make a pair and so do 2 and 4. Here 1 and 2 have a negative slope as their angles with positive x-axis are each obtuse, and by definition, slope of a line that makes α angle with positive x-axis is given by tan α, which turns out to be negative when 90° < α < 180°, and positive when 0° < α < 90°. Revise "Add Sugar To Coffee" concept of trigonometry for a better understanding.

In 1 and 2, 1 is n. In 3 and 4, 3 is n. In 1 and 3, 1 is n. In 2 and 4, 2 is n.

Wow.... had nearly forgotten this... was completely erased from my memory....

All -> 1st quadrant
Sin ->2nd quadrant
Tan->3rd quadrant
Cos->4th quadrant