If the distance between point R(a,a,a) and point J(6,-2,0) is 10, then the value of a could be?
Accepted Solution
A:
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)²Expanding the squares: 100 = 36 - 12a + a² + 4 + 2a + a² + a²Combining the constants: 100 - 36 - 4 = 60Combining the a terms: -12a + 2a = -10a;Combining the a² terms: 3a²Then the sum 100 = 36 - 12a + a² + 4 + 2a + a² + a² becomes 60 = -10a + 3a²Rewrite this in standard form for a quadratic: 3a² - 10a - 60 = 0Here 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·√205Then a = ------------------------- 6 5 ± √205Then a = ------------------------- 3