Linear Algebra Solutions 82

# Linear Algebra Solutions 82 - Let x = x1 y = y1 and...

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Let x = x 1 + α, y = y 1 + β and substitute in equation (4) to get 2( x 1 + α ) 2 +( y 1 + β ) 2 +3( x 1 + α )( y 1 + β ) - 5( x 1 + α ) - 4( y 1 + β )+3 = 0 (5) . Then equating the coefcients oF x 1 and y 1 to 0 gives 4 α + 3 β - 5 = 0 3 α + 2 β - 4 = 0 , which has the unique solution α = 2 , β = - 1. Then equation (5) simpli±es to 2 x 2 1 + y 2 1 + 3 x 1 y 1 = 0 = (2 x 1 + y 1 )( x 1 + y 1 ) . So relative to the x 1 , y 1 coordinates, equation (4) describes two lines: 2 x 1 + y 1 = 0 and x 1 + y 1 = 0. In terms oF the original x, y coordinates, these lines become 2( x - 2)+( y +1) = 0 and ( x - 2)+( y +1) = 0, i.e. 2 x + y - 3 = 0 and x + y - 1 = 0, which intersect in the point ( x, y ) = ( α, β ) = (2 , - 1) . (ii) Consider the equation 9 x 2 + y 2 - 6 xy + 6 x - 2 y + 1 = 0 . (6) Here Δ = f f f f f f 9 - 3 3 3 1 - 1
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