Answer:
Step-by-step 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
Answer:
Step-by-step 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
