Find the range of values of 'c' such that the two lines
x - y = 2 and cx + y = 3 intersect in the first quadrant.
All points in the first quadrant have both
their x and y coordinates positive.
x - y = 2
cx + y = 3
Adding the two equations term by term:
x + cx = 5
x(1+c) = 5
x = 5/(1+c)
This must be positive, so 1+c > 0 or c > -1
Substitute in the first:
x - y = 2
-y = 2 - x
y = -2 + x
y = -2 + 5/(1+c)
y = -2(1+c)/(1+c) + 5/(1+c)
y = (-2-2c)/(1+c) + 5/(1+c)
y = (-2-2c+5)/(1+c)
y = (3-2c)/(1+c)
This must be positive
This has critical numbers which are the
zeros of numerator and denominator.
Find the zero of the numerator:
3-2c=0
-2c=-3
c=3/2 = 1.5
Find the zero of the denominator:
1+c=0
c=-1
We place the critical numbers on a number line:
---------o---------o----------
-3 -2 -1 0 1 2 3 4
Choose a test point -2 in the leftmost interval
This is false so the interval
is not included in the range of values for c.
---------o---------o----------
-3 -2 -1 0 1 2 3 4
Choose a test point 0 in the middle interval
This is true so the interval
is included in the range of values for c.
---------o---------o----------
-3 -2 -1 0 1 2 3 4
Choose a test point 2 in the rightmost interval
This is false so the interval
is not included in the range of values for c.
Answer:
Edwin