find the equation of the straight line s through origin and at right angles to the lines x^2-5xy+4y^2=0

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SOLUTIONTO DETERMINEThe equation of the straight line s through origin and at right angles to the lines

[tex] \sf{ {x}^{2} – 5xy + 4 {y}^{2} = 0 }[/tex]

EVALUATIONHere the pair of lines are given by the equation

[tex] \sf{ {x}^{2} – 5xy + 4 {y}^{2} = 0 }[/tex]

We find the lines as below

[tex] \sf{ {x}^{2} – 5xy + 4 {y}^{2} = 0 }[/tex]

[tex] \sf{ \implies {x}^{2} – 4xy – xy + 4 {y}^{2} = 0 }[/tex]

[tex] \sf{ \implies x(x – 4y) – y(x – 4y) = 0 }[/tex]

[tex] \sf{ \implies (x – 4y) (x – y) = 0 }[/tex]

So given pair of lines are :

x – 4y = 0

x – y = 0

Now the equation of the line perpendicular to the line x – 4y = 0 and passing through the origin is 4x + y = 0

Again the equation of the line perpendicular to the line x – y = 0 and passing through the origin is x + y = 0

Hence the required equation is

[tex] \sf{(4x + y)(x + y) = 0}[/tex]

[tex] \sf{ \implies \: 4x (x + y) + y( x+ y)= 0}[/tex]

[tex] \sf{ \implies \: 4 {x}^{2} + 4xy + xy + {y}^{2} = 0}[/tex]

[tex] \sf{ \implies \: 4 {x}^{2} + 5xy + {y}^{2} = 0}[/tex]

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