If the pair of lines represented by 3x^2 - 5x | vedclass

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(D) The equation $6x^2 - xy - 5y^2 = 0$ can be factored as $(6x + 5y)(x - y) = 0$. This gives two lines: $y = x$ and $y = -\frac{6x}{5}$. Case $1$: If

Vedclass · Accepted Answer

$-\frac{17}{4}$

Vedclass · Answer

(D) The equation $6x^2 - xy - 5y^2 = 0$ can be factored as $(6x + 5y)(x - y) = 0$. This gives two lines: $y = x$ and $y = -\frac{6x}{5}$. Case $1$: If $y = x$ is a common line,substituting $y = x$ into $3x^2 - 5xy + Py^2 = 0$ gives $3x^2 - 5x^2 + Px^2 = 0$,which implies $x^2(P - 2) = 0$,so $P = 2$. Case $2$: If $y = -\frac{6x}{5}$ is a common line,substituting $y = -\frac{6x}{5}$ into $3x^2 - 5xy + Py^2 = 0$ gives $3x^2 - 5x(-\frac{6x}{5}) + P(-\frac{6x}{5})^2 = 0$. This simplifies to $3x^2 + 6x^2 + P(\frac{36x^2}{25}) = 0$,which is $9x^2 + \frac{36Px^2}{25} = 0$. Dividing by $9x^2$,we get $1 + \frac{4P}{25} = 0$,so $P = -\frac{25}{4}$. The sum of all possible values of $P$ is $2 + (-\frac{25}{4}) = \frac{8 - 25}{4} = -\frac{17}{4}$.

If the pair of lines represented by $3x^2 - 5xy + Py^2 = 0$ and $6x^2 - xy - 5y^2 = 0$ have one line in common,then the sum of all possible values of $P$ is

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