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x1+y*v1 = x2+y*v2
=>(x1-x2)+y(v1-v2) = 0
=>(x1-x2) = -y(v1-v2) Removing the '-ve' sign from RHS =>(x2-x1) = y(v1-v2)
=>(x2-x1)/(v1-v2) = y ----(1)
If you multiply -1 to both the numerator and denominator, then
=>(x1-x2)/(v2-v1) = y ----(2)
Thus equation (1) and (2) are the same.
Kangaroo
You are viewing a single comment's thread. Return to all comments →
x1+y*v1 = x2+y*v2
=>(x1-x2)+y(v1-v2) = 0
=>(x1-x2) = -y(v1-v2) Removing the '-ve' sign from RHS =>(x2-x1) = y(v1-v2)
=>(x2-x1)/(v1-v2) = y ----(1)
If you multiply -1 to both the numerator and denominator, then
=>(x1-x2)/(v2-v1) = y ----(2)
Thus equation (1) and (2) are the same.