I think you mean

In the XY-coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?

1) The product of the x-intercepts of line L and K is positive.

2) The product of the y-intercepts of line L and k is negative.

We know that the lines L and K intersect in Quadrant I (the top right quadrant).

The products of the slopes can be negative if the slopes are opposite signs. For examples, L = positive slope, K = negative slope, and vice versa.

1) The product of the x-intercepts being positive means that they both cross the x-axis BOTH where x < 0 , or where x > 0. But this does not imply that their slopes have to be the opposite signs of each other.

2) The product of the Y intercepts being negative means that one line crosses the y-axis above the line y=0, and the other line crosses the y-axis below the line where y=0. BUT, this does not imply that they have opposite sign slopes.

If both conditions were true, then the slopes MUST be opposite signs of each other.

K=a1x+b1
L=a2x+b2

is a1a2 < 0 ?



St1)

(-b1/a1) x (- b2/a2) > 0

or b1b2/a1a2 > 0

We can't decide whether a1a2 < 0 because we need to know the sign of b1b2

INSUF

St2)

b1b2 < 0 Nothing given for a1a2


INSUF


St1+St2)

b1b2/a1a2 > 0
b1b2 < 0

Hence, a1a2 < 0

Choose (C)

A straight line can be expreesed as :

x/a1+y/b1=1 (a1= X axis intercept, b1=Y axis intercept)

b1x+a1y=a1b1

y=-(b1/a1)x+a1b1 .......line K

similarly y=-(b2/a2)x+a2b2 .....line L

product of slopes = (b1b2/a1a2)

From (1) a1a2 >0 INSUFF
From (2) b1b2 >0 INSUFF

Combining product of slopes is positive

(C)

Bidisha800 wrote:A straight line can be expreesed as :

x/a1+y/b1=1 (a1= X axis intercept, b1=Y axis intercept)

b1x+a1y=a1b1

y=-(b1/a1)x+a1b1 .......line K

similarly y=-(b2/a2)x+a2b2 .....line L

product of slopes = (b1b2/a1a2)

From (1) a1a2 >0 INSUFF
From (2) b1b2 >0 INSUFF

Combining product of slopes is positive

(C)

Bringing tears to my eyes

logitech wrote:

Bidisha800 wrote:A straight line can be expreesed as :

x/a1+y/b1=1 (a1= X axis intercept, b1=Y axis intercept)

b1x+a1y=a1b1

y=-(b1/a1)x+a1b1 .......line K

similarly y=-(b2/a2)x+a2b2 .....line L

product of slopes = (b1b2/a1a2)

From (1) a1a2 >0 INSUFF
From (2) b1b2 >0 INSUFF

Combining product of slopes is positive

(C)

Bringing tears to my eyes

Here is a suggestion JACKA$$ - Wipe your last 3 sub 580 scores with your tears then rent a gun, buy a bullet and kill your self :twisted:

Am wrong or the "point (4,3)" information is just irrelevant ?

From (I) (bl*bx)/(ml*mk)>0 (INSUF), from (II) bk*bl<0 (INSUF), so with both (I) and (II) together you can answer if ml*mk<0. So the info about (4,3) is not needed.

Thanks,

yes, if you look at it from a purely 'equation of a line perspective' the point (4,3) info is not needed
product of slopes ( refer logitech's calculation) = b1b2/a1a2

to check if b1b2/a1a2 is < 0 we just need to know the signs of the numerator and the denominator and we are through!
Stmt 1 does just that but only provides sign of the denominator. so insufficient a1a2 >0

Stmt 2: provides sign of the numerator. numerator b1b2 < 0

only after combining the 2 statements can we see that b1b2/a1a2 < 0.
and that answers our question.

Drawing out each and every possibility on the graph is a tedious process.
I find the mathematical equivalent quite quick , if not simple.

a few general pointers
1. any linear equation of a line can be expressed as ax+by+c=0
ax+by = c
y-intercept is value of y when x =0 , i.e. y int= -c/b
x-intercept is value of x when y=0 , i.e. x int = -c/a
slope of a line passing through two points ( 0,-c/b) and (-c/a,0) is
(-c/b-0)/(0-(-c/a)) = - b/a
slope of a line = negative of y intercept / x intercept
its easier to remember this formula rather than its derivation.

FC wrote:Am wrong or the "point (4,3)" information is just irrelevant ?

From (I) (bl*bx)/(ml*mk)>0 (INSUF), from (II) bk*bl<0 (INSUF), so with both (I) and (II) together you can answer if ml*mk<0. So the info about (4,3) is not needed.

Thanks,

Hey!

(-c/b-0)/(0-(-c/a)) = - b/a

Should'nt the result of division be -a/b as -c/b*a/c=-a/b.

Appreciate your help!!!!

Bidisha800 ........... you are the meanest guy I have come acroos on any of the forums.. people like you should be banned from site for demeaning guys who are learning and looking for encoraugement .. shame....

Bidisha800 wrote:

logitech wrote:

Bidisha800 wrote:A straight line can be expreesed as :

x/a1+y/b1=1 (a1= X axis intercept, b1=Y axis intercept)

b1x+a1y=a1b1

y=-(b1/a1)x+a1b1 .......line K

similarly y=-(b2/a2)x+a2b2 .....line L

product of slopes = (b1b2/a1a2)

From (1) a1a2 >0 INSUFF
From (2) b1b2 >0 INSUFF

Combining product of slopes is positive

(C)

Bringing tears to my eyes

Here is a suggestion JACKA$$ - Wipe your last 3 sub 580 scores with your tears then rent a gun, buy a bullet and kill your self :twisted: