All help appreciated and 10 pts goes to best answer of course!
1. Show that the point (3,0,2) is equidistant from the points (1,-1,5) and (5,1,-1)
2. Find the perimeter of the triangle with vertices (-1,1,2), (2,0,3), and (3,4,5)
1. Show that the point (3,0,2) is equidistant from the points (1,-1,5) and (5,1,-1)
2. Find the perimeter of the triangle with vertices (-1,1,2), (2,0,3), and (3,4,5)
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just distance formula
(3,0,2) to (1,-1,5) = √(2² + 1² + 3²) = √14
(3,0,2) to (5,1,-1) = √(2² + 1² + 3²) = √14
(-1,1,2) to (2,0,3) is √(9+1+1) = √11
(2,0,3) to (3,4,5) is √(1+16+4) = √21
(3,4,5) to (-1,1,2) is √(16+9+9) = √34
so perimeter is √11 + √21 + √34 = 13.73
(3,0,2) to (1,-1,5) = √(2² + 1² + 3²) = √14
(3,0,2) to (5,1,-1) = √(2² + 1² + 3²) = √14
(-1,1,2) to (2,0,3) is √(9+1+1) = √11
(2,0,3) to (3,4,5) is √(1+16+4) = √21
(3,4,5) to (-1,1,2) is √(16+9+9) = √34
so perimeter is √11 + √21 + √34 = 13.73
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1. use the distance formula
d=√(x2-x1)²+(y2-y1)²+(z2-z1)²
d=√(1-3)²+(-1-0)²+(5-2)² =>d=√14
d=√(5-3)²+(1-0)²+(-1-2)² =>d=√14
the distances are congruent :)
2. again the same formula:
d= √(2-(-1))²+(0-1)²+(3-2)² =>d=√11
d= √(3-(-1))²+(4-1)²+(5-2)² =>d=√34
d=√(3-2)²+(4-0)²+(5-3)² =>d=√21
the perimeter is √11+√34+√21≈13.73
d=√(x2-x1)²+(y2-y1)²+(z2-z1)²
d=√(1-3)²+(-1-0)²+(5-2)² =>d=√14
d=√(5-3)²+(1-0)²+(-1-2)² =>d=√14
the distances are congruent :)
2. again the same formula:
d= √(2-(-1))²+(0-1)²+(3-2)² =>d=√11
d= √(3-(-1))²+(4-1)²+(5-2)² =>d=√34
d=√(3-2)²+(4-0)²+(5-3)² =>d=√21
the perimeter is √11+√34+√21≈13.73