Victoria V. answered • 09/28/18

Tutor

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Math Teacher: 20 Yrs Teaching/Tutoring Calculus, PreCalc, Alg 2, Trig

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!

Bader A.

09/28/18