The locus of the points which are at an equal | vedclass

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(B) The locus of points equidistant from two lines $L_1: a_1x + b_1y + c_1 = 0$ and $L_2: a_2x + b_2y + c_2 = 0$ is given by the angle bisectors: $\fr

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$99x + 77y - 133 = 0$

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(B) The locus of points equidistant from two lines $L_1: a_1x + b_1y + c_1 = 0$ and $L_2: a_2x + b_2y + c_2 = 0$ is given by the angle bisectors: $\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$.For the given lines $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$,the equations are $\frac{3x + 4y - 11}{5} = \pm \frac{12x + 5y + 2}{13}$.Case $1$ (Positive sign): $13(3x + 4y - 11) = 5(12x + 5y + 2) \implies 39x + 52y - 143 = 60x + 25y + 10 \implies 21x - 27y + 153 = 0$.Case $2$ (Negative sign): $13(3x + 4y - 11) = -5(12x + 5y + 2) \implies 39x + 52y - 143 = -60x - 25y - 10 \implies 99x + 77y - 133 = 0$.To check which line is near the origin,we test the origin $(0,0)$ in the expressions. For $3x + 4y - 11$,the value is $-11$. For $12x + 5y + 2$,the value is $2$. Since the signs are opposite,the origin lies between the lines. The bisector containing the origin is obtained by making the constant terms positive. The equation $99x + 77y - 133 = 0$ satisfies the condition.

The locus of the points which are at an equal distance from $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$ and which is near the origin is:

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