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(A) The equations of the angle bisectors of the lines $L_1: x + 2y - 11 = 0$ and $L_2: 3x - 6y - 5 = 0$ are given by $\frac{x + 2y - 11}{\sqrt{1^2 + 2

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$3x = 19$

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(A) The equations of the angle bisectors of the lines $L_1: x + 2y - 11 = 0$ and $L_2: 3x - 6y - 5 = 0$ are given by $\frac{x + 2y - 11}{\sqrt{1^2 + 2^2}} = \pm \frac{3x - 6y - 5}{\sqrt{3^2 + (-6)^2}}$.Simplifying,we get $\frac{x + 2y - 11}{\sqrt{5}} = \pm \frac{3x - 6y - 5}{3\sqrt{5}}$,which implies $3(x + 2y - 11) = \pm (3x - 6y - 5)$.Case $1$: $3x + 6y - 33 = 3x - 6y - 5 \implies 12y = 28 \implies 3y = 7$.Case $2$: $3x + 6y - 33 = -(3x - 6y - 5) \implies 3x + 6y - 33 = -3x + 6y + 5 \implies 6x = 38 \implies 3x = 19$.To find which bisector contains the point $(1, -3)$,we substitute $(1, -3)$ into the expressions $3y - 7$ and $3x - 19$.For $(1, -3)$,$3(-3) - 7 = -16$ and $3(1) - 19 = -16$.Checking the sign of the original lines at $(1, -3)$: $L_1(1, -3) = 1 + 2(-3) - 11 = -16$ and $L_2(1, -3) = 3(1) - 6(-3) - 5 = 16$.Since $L_1$ and $L_2$ have opposite signs at $(1, -3)$,we choose the bisector equation corresponding to the subtraction of the normalized forms,which yields $3x = 19$.

The equation of the bisector of the angle between the lines $x + 2y - 11 = 0$ and $3x - 6y - 5 = 0$ which contains the point $(1, -3)$ is

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