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Points A(1,2) and B(3,7) are two vertices of triangle ABC. The locus of the centroid G of triangle ABC is the line 2x - y = 0. What is the minimum distance from the locus of vertex C to the line 2x - y = 0?
- 0
- 1
- 2/sqrt(5)
- 1/sqrt(5)
Correct answer: 0
Solution
Let G = ((1+3+x3)/3, (2+7+y3)/3) = ((4+x3)/3, (9+y3)/3). Given 2*((4+x3)/3) - (9+y3)/3 = 0 => 2(4+x3) - (9+y3) = 0 => 8 + 2x3 - 9 - y3 = 0 => 2x3 - y3 = 1. Locus of C is the line 2x - y = 1. Distance from line 2x - y = 1 to line 2x - y = 0: These are parallel lines. Distance = |1 - 0| / sqrt(4+1) = 1/sqrt(5). The minimum distance from any point on the locus of C (line 2x-y=1) to the line 2x-y=0 is 1/sqrt(5).
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