Question

If the distance between point R(a,a,a) and point J(6,-2,0) is 10, then the value of a could be?

Answer 1

Explanation:

√(6-a)^2+(-2-a)^2+(0-a)^2 = 10

36-12a+a^2+4+4a+a^2+a^2 =100

3a^2 -8a -60=0

(3a+10)(a-6)=0

a= -3/10 or 6

Answer 2

Explanation:

Let's use the distance formula here.

The distance between R(a, a, a) and J(6, -2, 0) is 10. The whole formula is:

d = √(6 - a)^2 + (-2 - a)^2 + (0 - a)^2 = 10.

We must solve for a.

Note that d² = 100.

Thus, we have d² = 100 = (6 - a)² + (-2 - a)² + (-a)²

Expanding the squares:

100 = 36 - 12a + a² + 4 + 2a + a² + a²

Combining the constants:

100 - 36 - 4 = 60

Combining the a terms: -12a + 2a = -10a;

Combining the a² terms: 3a²

Then the sum 100 = 36 - 12a + a² + 4 + 2a + a² + a² becomes

60 = -10a + 3a²

Rewrite this in standard form for a quadratic:

3a² - 10a - 60 = 0

Here the coefficients are a = 3, b = -10 and c = -60, so the discriminant, b²-4ac is 100-4(3)(-60), or 100 + 720, or 820.

-(-10) ± √820)

Then a = -------------------------

2(3)

or:

10 ± √4·√205

Then a = -------------------------

6

5 ± √205

Then a = -------------------------

3