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(D) Given the equation of the pair of lines: $2x^2 - xy - 3y^2 + 6x + y + 4 = 0$.First,we factorize the homogeneous part $2x^2 - xy - 3y^2 = 0$:$2x^2

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$\frac{65}{\sqrt{26}}$

Vedclass · Answer

(D) Given the equation of the pair of lines: $2x^2 - xy - 3y^2 + 6x + y + 4 = 0$.First,we factorize the homogeneous part $2x^2 - xy - 3y^2 = 0$:$2x^2 - 3xy + 2xy - 3y^2 = 0$$x(2x - 3y) + y(2x - 3y) = 0$$(x + y)(2x - 3y) = 0$.Let the separate equations be $(x + y + m) = 0$ and $(2x - 3y + l) = 0$.Then,$(x + y + m)(2x - 3y + l) = 2x^2 - xy - 3y^2 + (l + 2m)x + (m - 3m)y + ml = 2x^2 - xy - 3y^2 + 6x + y + 4$.Comparing coefficients:$l + 2m = 6$ $(i)$$l - 3m = 1$ (ii)Subtracting (ii) from $(i)$: $5m = 5 \Rightarrow m = 1$.Substituting $m = 1$ into $(i)$: $l + 2 = 6 \Rightarrow l = 4$.The separate lines are $L_1: x + y + 1 = 0$ and $L_2: 2x - 3y + 4 = 0$.The perpendicular distance from $P(-1, 5)$ to $L_1$ is $d_1 = \frac{|-1 + 5 + 1|}{\sqrt{1^2 + 1^2}} = \frac{5}{\sqrt{2}}$.The perpendicular distance from $P(-1, 5)$ to $L_2$ is $d_2 = \frac{|2(-1) - 3(5) + 4|}{\sqrt{2^2 + (-3)^2}} = \frac{|-2 - 15 + 4|}{\sqrt{4 + 9}} = \frac{|-13|}{\sqrt{13}} = \sqrt{13}$.The product of the lengths is $d_1 	imes d_2 = \frac{5}{\sqrt{2}} 	imes \sqrt{13} = \frac{5\sqrt{13}}{\sqrt{2}} = \frac{5\sqrt{26}}{2}$.Wait,re-evaluating the calculation: $d_2 = \frac{13}{\sqrt{13}} = \sqrt{13}$. Product is $\frac{5\sqrt{13}}{\sqrt{2}} = \frac{5\sqrt{26}}{2}$.Checking the options,$\frac{65}{\sqrt{26}} = \frac{65\sqrt{26}}{26} = \frac{5\sqrt{26}}{2}$. Thus,the correct option is $D$.

Element	$M^{3+}/M$	$M^{+}/M$
$Al$	$-1.66$	$+0.55$
$Tl$	$+1.26$	$-0.34$

The product of the lengths of the perpendiculars drawn from the point $(-1, 5)$ to the pair of lines $2x^2 - xy - 3y^2 + 6x + y + 4 = 0$ is

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Element $M^{3+}/M$ $M^{+}/M$

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