Linked comprehension on straight lines

Question

The vertex $A$ of triangle $ABC$ is $(3,-1)$. The equations of the median $BE$ and the angular bisector $CF$ are $x-4y+10=0$ and $6x+10y-59=0$ respectively. Then

1:$\;\;\;$The equation of $AB$ must be

(A) $x+y=2$

(B)$18x+13y=41$

(C)$23x+y=70$

(D)$x+4y=0$

2: $\;\;\;$Slope of side $BC$ must be

(A)$\frac17$

(B)$\frac19$

(C)$\frac29$

(D)None of these

3:$\;\;\;$The length of the side $AC$ must be

(A)$\sqrt{83}$

(B)$\sqrt{85}$

(C)$\sqrt{71}$

(D)None of these

I am not able to approach this question. In the first question, it is clear that (C) and (D) are the wrong options, since they do not pass through vertex $A$. However, I am unable to find the other conditions for finding the equations.

Answer 1

The question may be wrong for the CF.

the approach is right for OP to eliminate (C) and (D). so (A) and (B) are all possible solutions. these two lines intersect $BF$, we get two possible $B_1,B_2$.

then we set $C (x_c,y_c) $, we have $AC$ cross $BF$ get $F$, sine $F$ is midpoint of $AC$, so we get $C$,then check $CB_1,CB_2$,we have two slopes ,with $CE$ is bisector , we can make sure which $B$ is right one. rest question is clear if $B$ is fixed.

but this question have no answer . see pic below:

you can see the slope of $CE$(sorry there is typo on the graphic) is always less both $AC,BC$, so it is impossible to be the bisector of $\angle BCA$.

Answer 2

My method is a long one. Mainly it relies on some observation skills. Also I am not sure whether the values are correct or not. So here goes:

Let $y=mx+c$ be the equation of AB. Since it passes through (3,-1), we must have$$-1=3m+c$$In the first option of first question,m=-1 and c=2 which satisfies the condition we just obtained. Hence equation of AB is x+y=2.

Next we go on to find the point of intersection of AB and FC. It comes out to be $P(\frac{79}4,\frac{-71}4)$. Let us denote C by (h,k). Clearly equation of the line passing though P and C is $$\frac{y+\frac{71}{4}}{\frac{-71}{4}-k}=\frac{x-\frac{79}{4}}{\frac{79}{4}-h}$$ Simplifying we get, $$y(\frac{79}{4}-h)+x(\frac{71}{4}+k)+c=0$$ Comparing it with the line 6x+10y-59=0, we get 2 equations, $$\frac{79}{4}-h=10$$ $$\frac{71}{4}+k=6$$ You will get the point C. Solve the equation of line AB with the median BE to get the point B. I think you can handle it from here.