Find the value of \alpha\(^2\) + \beta\(^2\) if \alpha + \beta = 2 and the distance between points (1, \alpha) and (\beta, 1) is 3 units
A
14
B
3
C
5
D
11
Verified Correct Answer: D.
Step-by-Step Mathematical Solution
PQ = \(\sqrt{(\beta - 1)^{2} + (1 - \alpha)^{2}}\)
3 = \(\sqrt{(\beta^{2} -2\beta^{2} + 1 + 1 - 2\alpha + \alpha^{2})}\)
3 = \(\sqrt{(\alpha^{2} + \beta^{2} - 2\alpha + 2\beta + 2)}\)
3 = \(\sqrt{(\alpha^{2} + \beta^{2} - 2(\alpha + \beta) + 2)}\)
3 = \(\sqrt{(\alpha^{2} + \beta^{2} - 2 * 2 + 2)}\)
3 = \(\sqrt{(\alpha^{2} + \beta^{2} - 2)}\)
9 = \(\alpha^{2} + \beta^{2} - 2)\)
\(\alpha^{2} + \beta^{2}\) = 9 + 2
\(\alpha^{2} + \beta^{2}\) = 11
Source: Myschool Past Questions Archive (Record #7958)