Find a if Q(a,2a) is square root of 5 units from R(-1,1)?

Question

please help. i suck at maths

Answer 1

Just use the distance formula:

d = Sqrt[ (x2 - x1)^2 + (y2-y1)^2 ]

Substituting:

Sqrt[5] = Sqrt[ (a+1)^2 + (2a-1)^2 ]

5 = a^2 + 2a + 1 + 4a^2 -4a + 1

5a^2 - 2a - 3 = 0

Solving the quadratic:

(5a + 3)(a - 1) = 0

a = 1 or a = -3/5

Now just sub into Q(a, 2a) to find your possible points.

Answer 2

Use the distance formula of the two points given:

√[(-1 - a)² + (1 - 2a)²]

However, we already know the distance, which is √5 units so:

√[(-1 - a)² + (1 - 2a)²] = √5

Solve this using simple algebra - first, square both sides:

(-1 - a)² + (1 - 2a)² = 5

Expand the brackets using FOIL:

1 + a² + 2a + 1 + 4a² - 4a = 5

Collect like-terms (simplify):

5a² - 2a + 2 = 5

5a² - 2a - 3 = 0

Use the quadratic formula:

a = [-B ± √(B² - 4AC)] / 2A

a = [2 ± √((-2)² - 4*5*-3)] / 2*5

a = [2 ± √(4 + 60)] / 10

a = (2 ± √64) / 10

a = (2 ± 8) / 10

a = (1 ± 4) / 5

a = 1

a = -3/5

There are 2 solutions to a as it is a quadratic equation.