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(D) The given pair of lines is $8x^2 - 14xy + 5y^2 = 0$. To find the individual lines,we factorize the quadratic equation: $8x^2 - 10xy - 4xy + 5y^2 =

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$p = q$

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(D) The given pair of lines is $8x^2 - 14xy + 5y^2 = 0$. To find the individual lines,we factorize the quadratic equation: $8x^2 - 10xy - 4xy + 5y^2 = 0$ $2x(4x - 5y) - y(4x - 5y) = 0$ $(2x - y)(4x - 5y) = 0$. So,the two lines are $L_2: 2x - y = 0$ and $L_3: 4x - 5y = 0$. The third line is $L_1: x - 2y + 3 = 0$. The intersection of $L_2$ and $L_3$ is the origin $A(0, 0)$. For $L_1$ and $L_2$: $x = 2y$,substituting into $L_1$ gives $2y - 2y + 3 = 0$,which is impossible. Wait,let us re-solve: Intersection of $L_1$ and $L_2$ $(x = 2y)$: $2y - 2y + 3 = 0$ is not possible. Let's re-check the intersection of $L_1$ and $L_3$ $(4x = 5y)$: $x - 2(4x/5) + 3 = 0$ $\Rightarrow x - 8x/5 = -3$ $\Rightarrow -3x/5 = -3$ $\Rightarrow x = 5$. Then $y = 4$. So $C(5, 4)$. Intersection of $L_1$ and $L_2$ $(y = 2x)$: $x - 2(2x) + 3 = 0$ $\Rightarrow -3x = -3$ $\Rightarrow x = 1$. Then $y = 2$. So $B(1, 2)$. The vertices of the triangle are $A(0, 0)$,$B(1, 2)$,and $C(5, 4)$. The centroid $(p, q)$ is given by $(\frac{0+1+5}{3}, \frac{0+2+4}{3}) = (2, 2)$. Since $p = 2$ and $q = 2$,we have $p = q$.

If $(p, q)$ is the centroid of the triangle formed by the lines $8x^2 - 14xy + 5y^2 = 0$ and $x - 2y + 3 = 0$,then

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