How can all the options be correct in this question?

Question

Question: The equations of two sides $AB$ and $AC$ of an isosceles triangle $ABC$ are $x+y=5$ and $7x-y=3$ respectively. If the area of the Triangle $ABC$ is 5 sq. units , then the possible equations of the side $BC$ is (are): ?

$A). x-3y+1=0$
$B). 3x+y+2=0$
$C). x-3y-21=0$
$D). 3x+y-12=0$

Answer given :$A,B,C,D$

Making an attempt: Obviously, $A$ must be $(1,4)$. Now after that, if I consider a point $(x_1,y_1)$ on $AB$ and $(x_2,y_2)$ on $BC$, I could see that the area should be :

$$\frac{1}{2} \begin{bmatrix}1&4&1\\x_1&y_1&1\\x_2&y_2&1 \end{bmatrix}=5$$ (treat it as a determinant please)

And, I assumed that $AB=BC$ which gave me:

$$(2-x_1-x_2)(x_2-x_1)=(8-y_1-y_2)(y_1-y_2)$$

But sadly none of the equations could satisfy these conditions. True enough, my assumption can be wrong but I can't seem to make any headway in this question. Please Guide

Answer 1

WLOG, $B(a,5-a);C(b,7b-3)$

We have $$(a-1)^2+(5-a-4)^2=(b-1)^2+(7b-3-4)^2$$

$$\iff2(a-1)^2=50(b-1)^2\implies a-1=\pm5(b-1)$$

For $a-1=5(b-1), a=5b-4$

$\implies|BC|^2=(a-b)^2+(5-a-7b+3)^2=(5b-4-b)^2+\{8-7b-(5b-4)\}^2=(b-1)^2(4^2+12^2)$

Now if $M$ is the midpoint of $BC,M:\left(\dfrac{a+b}2,\dfrac{5-a+7b-3}2\right)$

Replace $a$ with $5b-4$

Now $|AM|$ is also the perpendicular distance of $BC$ from $A$

We have $$\dfrac{|AM|\cdot|BC|}2=5$$

Similarly for $a-1=-5(b-1)$

Answer 2

Using Theorem: In an isosceles triangle the bisector of the vertex angle cuts the opposite side in half.

Now the equation of $AM$ where $M$ is the midpoint of $BC$

$$\dfrac{x+y-5}{\sqrt{1^2+1^2}}=\pm\dfrac{7x-y-3}{\sqrt{7^2+1^2}}$$

$$\iff5(x+y-5)=\pm(7x-y-3)$$

From the other answer, $M(3b-2,6-b)$

So, we can find $b$

Also, use $AM\perp BC$