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. Ex press your answer as the ordered pair (a, b)​

The slope of the line passing through points A and B is

slope = (b ² - a ²) / (b - a) = ((b - a) (b + a)) / (b - a) = b + a

Since this line passes through (0, 3), its equation would be

y - 3 = (b + a) (x - 0)   ==>   y = (b + a) x + 3

It also passes through (-3/2, 0), so that

y - 0 = (b + a) (x + 3/2)   ==>   y = (b + a) x + 3/2 (b + a)

If these equations describe the same line, then they must have the same slope and y-intercept, so that

3 = 3/2 (b + a)   ==>   a + b = 2

It also passes through (a, a ²), so that

a ² = (b + a) a + 3   ==>   a ² = ab + a ² + 3   ==>   ab = -3

Solving for b in the first underlined equation, we get

b = 2 - a

Substituting into the second equation and solving for a gives

a (2 - a) = -3   ==>   2a - a ² = -3   ==>   a ² - 2a - 3 = (a - 3) (a + 1) = 0

==>   a = 3   or   a = -1

==>   b = -1   or   b = 3

Since a < b, we pick a = -1 and b = 3, so the solution is (a, b) = (-1, 3).

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step-by-step explanation:

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.