The lines {a^2}{x^2} + bc{y^2} = a(b + c)xy w | vedclass

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(A) The general equation of a pair of straight lines is given by $Ax^2 + 2Hxy + By^2 = 0$.Comparing the given equation ${a^2}{x^2} - a(b + c)xy + bc{y

Vedclass · Accepted Answer

$a = 0$ or $b = c$

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(A) The general equation of a pair of straight lines is given by $Ax^2 + 2Hxy + By^2 = 0$.Comparing the given equation ${a^2}{x^2} - a(b + c)xy + bc{y^2} = 0$ with the standard form,we have $A = a^2$,$2H = -a(b + c)$,and $B = bc$.The lines are coincident if $H^2 - AB = 0$.Substituting the values,we get $\left\{ -\frac{a(b + c)}{2} \right\}^2 - (a^2)(bc) = 0$.$\frac{a^2(b + c)^2}{4} - a^2bc = 0$.$a^2(b^2 + 2bc + c^2) - 4a^2bc = 0$.$a^2(b^2 - 2bc + c^2) = 0$.$a^2(b - c)^2 = 0$.Thus,$a = 0$ or $b = c$.

The lines ${a^2}{x^2} + bc{y^2} = a(b + c)xy$ will be coincident,if

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