Let a and b be real numbers, where a<b, and let A= (a,a^2) and B=(b,b^2). The line AB (meaning the unique line that contains the point A and the point B) has x-intercept (-3/2,0) and y-intercept (0,3). Find a and b.

Answer 1

Are we to assume that your line AB is a straight line?
If so, we can find the slope of AB
slope AB = (b^2 -a^2)/(b-a)
= (b-a)(b+a)/(b-a)
= b+a

we can write the equation as
y - a^2 = (b+a)(x-a)

also (3/2,0) lies on it, so
0-a^2 = (b+a)(-3/2 - a)
-a^2 = -3b/2 -ab - 3a/2 - a^2
3b/2 - 3a/2 = -ab
3a + 3b = -2ab , #1

and (0,3) lies on it
3 - a^2 = (b+a)(-a)
3-a^2 = -ab - a^2
then ab = -3 or a = -3/b

sub that into #1
3(-3/b) + 3b = -2(-3)
-3/b + b = 2
times b
-3 + b^2 = 2b
b^2 - 2b - 3 = 0
(b-3)(b+1) = 0
b = 3 or b = -1

if b = 3 , a = -3/3 = -1
if b = -1 , a = -3/-1 = 3 , but a should be < b

so a = -1 and b = 3

Answer 2

-1,3

Answer 3

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