find p,if distance between (4,5,p) & (7,1,-13) is 13.
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distance between two points: (x1,y1,z1) & (x2,y2,z2) :
= √[ (x2-x1)²+ (y2-y1)² + (z2-z1)² ]
So: 13 = √ [ (7-4)² + (1-5)² + (-13-p)² ]
13² = 9+16 + 13² + 26 p + p²
p² + 26 p + 25 = 0
p = -13 +- 12 = -25 or -1
= √[ (x2-x1)²+ (y2-y1)² + (z2-z1)² ]
So: 13 = √ [ (7-4)² + (1-5)² + (-13-p)² ]
13² = 9+16 + 13² + 26 p + p²
p² + 26 p + 25 = 0
p = -13 +- 12 = -25 or -1
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