If p_1 and p_2 denote the lengths of perpendi | vedclass

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(C) The given equation of the pair of lines is $15 x^2 + 31 x y + 14 y^2 = 0$. Factoring the quadratic expression: $15 x^2 + 10 x y + 21 x y + 14 y^2

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$2$

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(C) The given equation of the pair of lines is $15 x^2 + 31 x y + 14 y^2 = 0$. Factoring the quadratic expression: $15 x^2 + 10 x y + 21 x y + 14 y^2 = 0$ $\Rightarrow 5 x(3 x + 2 y) + 7 y(3 x + 2 y) = 0$ $\Rightarrow (3 x + 2 y)(5 x + 7 y) = 0$. Thus,the two lines are $L_1: 3 x + 2 y = 0$ and $L_2: 5 x + 7 y = 0$. The perpendicular distance from $(2,3)$ to $3 x + 2 y = 0$ is $d_1 = \frac{|3(2) + 2(3)|}{\sqrt{3^2 + 2^2}} = \frac{12}{\sqrt{13}}$. The perpendicular distance from $(2,3)$ to $5 x + 7 y = 0$ is $d_2 = \frac{|5(2) + 7(3)|}{\sqrt{5^2 + 7^2}} = \frac{31}{\sqrt{74}}$. Since $p_1 > p_2$,we have $p_1 = \frac{31}{\sqrt{74}}$ and $p_2 = \frac{12}{\sqrt{13}}$. Now,calculating $p_1^2 + \frac{1}{74} - p_2^2 + \frac{1}{13} = \frac{31^2}{74} + \frac{1}{74} - \frac{12^2}{13} + \frac{1}{13} = \frac{961+1}{74} - \frac{144-1}{13} = \frac{962}{74} - \frac{143}{13} = 13 - 11 = 2$.

If $p_1$ and $p_2$ denote the lengths of perpendiculars from $(2,3)$ onto the lines given by $15 x^2+31 x y+14 y^2=0$,and if $p_1 > p_2$,then $p_1^2 + \frac{1}{74} - p_2^2 + \frac{1}{13}$ is equal to

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