[Solced] Consider the following equations of straight lines

Previously Asked in : Gate CE-afternoon season 2022 ||

L4 is perpendicular to L2 and L4 is parallel to L3

L1 is parallel to L2 and L1 is perpendicular to L3

L2 is parallel to L4 and L2 is perpendicular to L1

L3 is perpendicular to L4 and L3 is parallel to L2

Answer (Detailed Solution Below)

Option L₄ is parallel to L₂ and L₄ is parallel to L₃ is correct

Detailed Solution :

Line $L_1 : 2x-3y = 5\\ y = \dfrac {4x-5} {3}$

$\therefore \, \dfrac {dy} {dx} \, = \, \dfrac {2} {3}$

Similarly for other lines

$\textrm {Line} \, L_2 : 3x + 2y = 8 \\ \dfrac {dy} {dx} \, = \, \dfrac {-3} {2} \\ \textrm{Line} \, L_3 : 4x -6y = 5 \\ \dfrac {dy} {dx} \, = \, \dfrac{2} {3} \\ \textrm {Line} \, L_4 : 6x -9y = 6 \\ \dfrac {dy} {dx} \, = \, \dfrac{2} {3}$

\therefore \quad \left (\dfrac{dy}{dx}\right)_{L_1} = \, \left (\dfrac{dy} {dx} \right)_{L_3} = \, \left (\dfrac {dy}{dx} \right)_{L_4} = \, \dfrac {2}{3}

and $\left(\frac{dy}{dx} \right)_{L_2} \times \left(\frac{dy}{dx}\right)_{L_1} = \left(\frac{dy}{dx} \right)_{L_2} \times \left( \frac{dy} {dx} \right)_{L_3} = \left (\frac{dy} {dx} \right)_{L_2} \times \left (\frac{dy}{dx} \right)_{L_4} = -1$

$\therefore$ L₁, L₃, L₄ are parallel to each other and perpendicular to L₂ .

Correct option is: L₄ is parallel to L₂ and L₄ is parallel to L₃

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