Equations of the sides of the triangle having $(3,-1)$ as a vertex, $x - 4y + 10 = 0$ and $6x + 10y - 59 = 0$ being the equations of an angle bisector and a median respectively drawn from different vertices. The equation of one side is:
Step-by-Step Solution
Key Concept: The midpoint of one side lying on the median from another vertex constrains vertex positions, combined with angle bisector slope relationships.
Given vertex $A = (3, -1)$, median through $B$ is $6x + 10y - 59 = 0$, and angle bisector through $C$ is $x - 4y + 10 = 0$. The midpoint $D$ of $AC$ lies on the median through $B$, giving $C = (10, 5)$. From $AC$ equation $6x - 7y = 25$ and using the angle bisector condition that $BC$ and $AC$ are equally inclined to $x - 4y + 10 = 0$, we find the slope of $BC$ is $m = -2/9$. This gives equation of $BC$ as $2x + 9y = 65$. Solving simultaneously with the median equation yields $B = (-7/2, 8)$ and equation of $AB$ is $18x + 13y = 41$.
Correct Answer: 1,2,3