Please help. Thanks

Question

Given the coordinates of points A (2,-3) and B (x,3x-9), and the distance AB= square root of 40. Find 2 possible coordinates of point B.

Answer 1

The distance formula is d = √(X₂ - X₁)² + (Y₂ - Y₁)²

Replace your coordinates and solve for each part under the radical.

(2 - X)² = X² - 4X + 4

[-3 - (3X - 9)]² = 9(X² - 4X + 4

This means there are a total of 10 of them altogether: 10(X² - 4x + 4)

The problem tells us that the distance = the sqrt of 40, so √40 = √10(X² - 4x + 4)

Simplify both sides by pulling out the square root on each side: 2√10 = (x - 2)√10

Since we are solving for X, we can eliminate the √10 on both sides. Now all we have is 2 = (X - 2).

This means X = 4

If the B coordinate = (X, 3X - 9), then one possible set of values for these coordinates is (4, 3), since (3*4) - 9 = 3.

Since you are given the set distance of √40 and point A (2,-3), point B can really be in any direction on the graph from point A as long as it is √40 away.

If you use a compass and stick it at point A in your paper (2,-3), point B can be any of the points around the circle you draw that is √40 (or 2√10) distance away from point A. If you do this with graph paper, this is a really good way to check your answers (or estimate them ahead of time) because you will have a good idea if you're solving it correctly.

Answer 2