Find the equation of the line which is equidi | vedclass

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Question

(A) The given lines are $9x + 6y - 7 = 0$ and $3x + 2y + 6 = 0$.To make the coefficients of $x$ and $y$ the same,multiply the second equation by $3$:$

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$18x + 12y + 11 = 0$

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(A) The given lines are $9x + 6y - 7 = 0$ and $3x + 2y + 6 = 0$.To make the coefficients of $x$ and $y$ the same,multiply the second equation by $3$:$3(3x + 2y + 6) = 3(0) \Rightarrow 9x + 6y + 18 = 0$.Now the lines are $L_1: 9x + 6y - 7 = 0$ and $L_2: 9x + 6y + 18 = 0$.The line equidistant from two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $ax + by + \frac{c_1 + c_2}{2} = 0$.Here,$a = 9, b = 6, c_1 = -7, c_2 = 18$.The required equation is $9x + 6y + \frac{-7 + 18}{2} = 0$.$9x + 6y + \frac{11}{2} = 0$.Multiplying by $2$,we get $18x + 12y + 11 = 0$.

Find the equation of the line which is equidistant from the parallel lines $9x + 6y - 7 = 0$ and $3x + 2y + 6 = 0$.

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