Pre calculus question !!!

Question

Given two lines   L₁：x + ay + b = 0   and   L₂：x + cy + d = 0   as detailed below :

details:   1.  L₁ and L₂ cross each other in the section 1 of the Cartesian coordinate system.
2.  L₁ crosses the positive section of X coordinate and crosses the negative section of Y coordinate. 
3.  L₂ crosses the positive section of the Y coordinate and the negative section of X coordinate.

How many among the real numbers a, b, c, d, a–c are positive ??

 

(a) 1    (b) 2    (c) 3    (d) 4    (e) 5   

Answer 1

Have you graphed this?  You really need to graph this to see what is going on.  Sorry I cannot draw it for an answer.

 

Notice they intersect in the 1st quadrant.

For L1 to cross the pos x-axis and the neg y-axis means it has a positive slope.  In slope intercept form, this would mean that for y=mx+b,  the m MUST BE POSITIVE.

For L2 to cross the neg x-axis and the pos y-axis means it also has a positive slope.  

 

If you rearange each linear equation to be in the form of  y = mx + b,  you get

 

L1:  y = (-1/a)x - (b/a)   So the slope is (-1/a) which must be positive, so a must be negative.

 

L2:  y = (-1/c)x - (d/c)    So the slope is (-1/c) which must be positive, so c must be negative.

 

 

The final term  of   y=mx+b, the b is the y-intercept.  

 

L1 crosses the negative y-axis, so -(b/a) must be negative.  a is negative, so be must also be negative so that -(b/a) is negative.

 

L2 crosses the positive y-axis, so -(d/c) must be positive, but we already know that c is negative, so d must be positive.

 

As for a - c:  I graphed it and used real numbers and found that a - c = positive.

 

So 

 

a = neg

b = neg

c = neg

d = pos

a-c = pos

 

A graph is REALLY HELPFUL for  this problem!