Two equal sides $AB$ and $AC$ of an acute angle triangle $ABC$ are formed by the equations $7x - y + 3 = 0$ and $x + y - 3 = 0$, side $BC$ is $ax + by - 31 = 0$ where $a - 10b = 31$. Value of $2a + 10b + 31 = 0$ is ____.
Step-by-Step Solution
Key Concept: The angle bisector formula uses the signed ratio of distance formulas; the perpendicular to the bisector gives the opposite side of the triangle.
Lines $L_1: 7x - y + 3 = 0$ and $L_2: -x + y + 3 = 0$ form a triangle with point $A$ at the apex. Since $a_1a_2 + b_1b_2 = (7)(-1) + (-1)(1) = -8 < 0$, the angle bisector with positive sign is: $\frac{7x-y+3}{\sqrt{50}} = \frac{-x+y+3}{\sqrt{2}}$, simplifying to $3x + y - 3 = 0$. Line $BC$ passes through $(1, -10)$ with slope $-\frac{1}{m_{AD}} = \frac{1}{3}$, giving equation $x - 3y - 31 = 0$. With $a=1, b=-3, c=-31$: $2a + 10b - c = 2 - 30 + 31 = 3$.
Correct Answer: 3