Let the lines be L_1: 2x + 3y - 7 = 0 and L_2 | vedclass

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(D) The given lines are $L_1: 2x + 3y - 7 = 0$ and $L_2: 2x + 3y - 12 = 0$.For point $A(3, -5)$,evaluate the expressions:$L_1(3, -5) = 2(3) + 3(-5) -

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None of these.

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(D) The given lines are $L_1: 2x + 3y - 7 = 0$ and $L_2: 2x + 3y - 12 = 0$.For point $A(3, -5)$,evaluate the expressions:$L_1(3, -5) = 2(3) + 3(-5) - 7 = 6 - 15 - 7 = -16 $L_2(3, -5) = 2(3) + 3(-5) - 12 = 6 - 15 - 12 = -21 Since both values have the same sign,point $A$ does not lie between the lines.Perpendicular distance $d_1$ from $A$ to $L_1$ is $\frac{|2(3) + 3(-5) - 7|}{\sqrt{2^2 + 3^2}} = \frac{|-16|}{\sqrt{13}} = \frac{16}{\sqrt{13}}$.Perpendicular distance $d_2$ from $A$ to $L_2$ is $\frac{|2(3) + 3(-5) - 12|}{\sqrt{2^2 + 3^2}} = \frac{|-21|}{\sqrt{13}} = \frac{21}{\sqrt{13}}$.The distance between the parallel lines is $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|-7 - (-12)|}{\sqrt{2^2 + 3^2}} = \frac{5}{\sqrt{13}}$.Thus,none of the given options $A, B, C$ are correct.

Let the lines be $L_1: 2x + 3y - 7 = 0$ and $L_2: 2x + 3y - 12 = 0$. For the point $A(3, -5)$,which of the following is true?

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