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Eliminate l,m between the equations ${\text{lx + my = a, mx - ly = b, }}{{\text{l}}^2}{\text{ + }}{{\text{m}}^2}{\text{ = 1}}$ .

Last updated date: 16th Jul 2024
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Hint:- Square both the equations given in terms of x & y and add.

Let the given equation be
${\text{lx + my = a }} \cdots \left( 1 \right) \\ {\text{mx - ly = b }} \cdots \left( 2 \right) \\ {{\text{l}}^2}{\text{ + }}{{\text{m}}^2}{\text{ = 1 }} \cdots \left( 3 \right) \\$
For eliminating l and m, we need to perform squaring on both sides of equation (1) and (2).
${\left( {{\text{lx + my}}} \right)^2} = {\text{ }}{{\text{a}}^2}{\text{ }} \cdots \left( 4 \right) \\ {\left( {{\text{mx - ly}}} \right)^2}{\text{ = }}{{\text{b}}^2}{\text{ }} \cdots \left( 5 \right) \\$
On adding equations (4) and (5) and simplifying the equation, we get
${\left( {{\text{lx + my}}} \right)^2}{\text{ + }}{\left( {{\text{mx - ly}}} \right)^2}{\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ }} \cdots \left( 6 \right) \\ {{\text{l}}^2}{{\text{x}}^2}{\text{ + }}{{\text{m}}^2}{{\text{y}}^2}{\text{ + 2(lx)(my) + }}{{\text{m}}^2}{{\text{x}}^2}{\text{ + }}{{\text{l}}^2}{{\text{y}}^2}{\text{ - 2(mx)(ly) = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2} \\$
Cancelling (2lmxy) terms and taking ${{\text{l}}^2}{\text{ + }}{{\text{m}}^2}$ common , we get
$\left( {{{\text{l}}^2}{\text{ + }}{{\text{m}}^2}} \right){{\text{x}}^2}{\text{ + }}\left( {{{\text{l}}^2}{\text{ + }}{{\text{m}}^2}} \right){{\text{y}}^2} = {\text{ }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ }} \cdots \left( 7 \right)$
Putting the value of ${{\text{l}}^2}{\text{ + }}{{\text{m}}^2}$ from equation (3), we get
${{\text{x}}^2}{\text{ + }}{{\text{y}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}$