The ratio in which the line 3x + 4y + 2 = 0 d | vedclass

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(B) The given lines are $L_1: 3x + 4y + 5 = 0$,$L_2: 3x + 4y - 5 = 0$,and the dividing line $L: 3x + 4y + 2 = 0$.The distance $d_1$ between $L$ and $L

Vedclass · Accepted Answer

$3:7$

Vedclass · Answer

(B) The given lines are $L_1: 3x + 4y + 5 = 0$,$L_2: 3x + 4y - 5 = 0$,and the dividing line $L: 3x + 4y + 2 = 0$.The distance $d_1$ between $L$ and $L_1$ is given by $d_1 = \left| \frac{2 - 5}{\sqrt{3^2 + 4^2}} \right| = \frac{|-3|}{5} = \frac{3}{5}$.The distance $d_2$ between $L$ and $L_2$ is given by $d_2 = \left| \frac{2 - (-5)}{\sqrt{3^2 + 4^2}} \right| = \frac{|7|}{5} = \frac{7}{5}$.The ratio in which the line $L$ divides the distance between $L_1$ and $L_2$ is $d_1 : d_2 = \frac{3}{5} : \frac{7}{5} = 3:7$.

The ratio in which the line $3x + 4y + 2 = 0$ divides the distance between $3x + 4y + 5 = 0$ and $3x + 4y - 5 = 0$ is:

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