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1 Comment Already

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Mohasa Said,
January 17th, 2011 @1:23 am  

Plane: 2x+4y+3z=16

The x-intercept occurs where y = 0 and z = 0.

Solve for x: 2x + 0 + 0 =16 or x = 8

So P1 is at (8, 0, 0)

The y-intercept occurs where x = 0 and z=0.

Solve for y: 4y = 16 or y = 4

Point P2 is (0, 4, 0)

The distance between points P1 and P2 is given by:

d = ?{(0 – 8)² + (4 – 0)² +(0 – 0)²}

or d = ?(64 + 16) = ?80

d = 4?5 <======Ans

Hope this helps

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