If the slopes of the line l1 and l2 are of the same sign, is

many thanks +1

however slope of m1>m2 and not m1<m2 as you have arrived at, guess small oversight in your calculation.

Here we go:


\(b=m_1.a+c_1\), and \(b=m_2.a+c_2\) these are the two equations of the line from statement 1

\(c_1= b-m_1a\) and \(c_2=b-m_2a\).. ( 1)

x intercept's \(\frac{-c_1}{m_1}\) and \(\frac{-c_2}{m_2}\)...(2)

given \(\frac{-c_1}{m_1} > \frac{-c_2}{m_2}\). from statement 2

now substituting the values of \(c_1\) and \(c_2\) from ..(1) in ...(2)

\(\frac{-(b-m_1a)}{m_1} > \frac{-(b-m_2a)}{m_2}\)

\(\frac{-b+m_1a}{m_1} > \frac{-b+m_2a}{m_2}\) ,multiplying the negative sign in the numerator

\(\frac{-b}{m_1}+a >\frac{-b}{m_2}+a\) , dividing LHS by \(m_1\) and RHS by \(m_2\)
\(\frac{-b}{m_1}>\frac{-b}{m_2}\), cancelling out a from both sides

\(\frac{b}{m_1}<\frac{b}{m_2}\), Multiplying both sides by negative and reversing the > sign to < , since in inequality we reverse the operator when we multiply( or divide)by a negative sign. ( Guess you forgot this part )

so finally we are left with \(\frac{b}{m_1}<\frac{b}{m_2}\)

\(\frac{1}{m_1}<\frac{1}{m_2}\) , cancelling b from both sides

\(\frac{1}{m_1}<\frac{1}{m_2}\)

Now we are still not sure of the signs of slope , both could be + or both could be -

if both are positive then
\(m_1>m_2\)

if both are negative

then we have \(\frac{1}{- m_1}<\frac{1}{- m_2}\)

here also we have \(m_1>m_2\)

hence \(m_1>m_2\) for all cases and hence sufficient

Hope I have done everything correctly


Thank you for showing the way.

My approach

y = mx + f (Line 1)

y = nx + g (Line 2)

Is m>n or m-n>0?

(1) b=am + f

b = an + g

m-n = g-f/a where 'a' is a positive integer

So m>n only when g>f Insuff

(2) n=-f/x
m=-g/x

-f>-g
g>f

Insuff

(1) and (2)

We know that g>f therefore m>n

C

L2 is steeper thus its slope is more negative (the absolute value of its slope is greater) than the slope of L2 ... Does it mean than L1??

Yes, that is correct.

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Hi can anyone help me here, in the figure above for the blue case it looks to me that L2 is steeper than L1, lets take a very large value and a very small value for the x intercept.Lets suppose L1 passes through 100 and L2 passes through 5,so x intercept of L1 is more than L2,since L2 is closer to the y axis on account of lesser x intercept it is more vertical hence its steeper and hence it looks to me that for Blue case( when both the lines have negative slope ) slope of L2 is greater than L1.

Also Visually in the Graph above for the Blue case it looks as though L2 is steeper than L1, what am I missing here?

ALTERNATE SOLUTION:

Equation 1 : l1 : y=(m1)X+c1 , where m1 is slope and c1 is constant
Equation 2 : l2 : y=(m2)X+c2 , where m2 is slope and c2 is constant

Also given that slope are of same sign. Means--either both m1, m2 are positive or both are negative.

Statement 1 : intersect at point (a,b) where in both a and b are positive.

(1)==> y=b=(m1)a+c1=(m2)a+c2 -simply putting coordinates in both the equations of lines.

But from here we cannot conclude whether m1 > or < m2

Hence Insufficient.


Statement 2 : The X-intercept of line l1 is greater than the X-intercept of line l2

==> As value of x, when y=0, is the X intercept. Thus, X intercept for l1 and l2 are:
l1: -c1/(m1)
l2: -c2/(m2)

(2)==> -c1/(m1) > -c2/(m2), Since we dont know the sogn of c1 and c2 and we dont know there values we cannot conclude anything from this.

Hence Insufficient.


Combining both the statements :
and solving (1) and (2) we get,

a-b/(m1) >a-b/(m2) ---> m1<m2 Hence SUFFICIENT. Thus answer is C.

Hi

I have been able to figure out the graphical way by myself, but I cannot figure out how you combined the two equations to get \(\frac{(a-b)}{m1} >\frac{(a-b)}{m2}\) in your algebraic method, can anyone help with the algebraic method here.

Thanks

Hi Bunuel, could you kindly enlighten on 2 questions please?
1 - How do we know the slopes of the lines are of the same sign as per the highlighted quote above?
2 - Why is that information on X-intercept can be used to solve this question, but not on another question when the intersection point of 2 lines is (-2,4).
This "other" question is referenced below, posted by dreambeliever

Other question: Lines m and n lie in the xy-plane and intersect at the point (-2; 4). Is the slope of line m less than the slope of line n?

(1) The x-intercept of line m is greater than the x-intercept of line n.
(2) The y-intercept of line n is greater than the y-intercept of line m.

PS: Apologies for any errors in posting, this is my first post.

Thank you!

Hi Bunuel, this distinction between 'greater/lesser' and 'steeper/less steep' still eludes me.

So basically for the red lines with POSITIVE slope, we say that the slope of line 1 is GREATER because it is STEEPER.
But, for the blue lines with NEGATIVE slope, we say that the slope of line 1 is SMALLER because it is LESS STEEP

How would you compare the slope of red line1 with blue line2? Which one has a 'greater' slope?

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