$ABC$ is a triangle and $X$ is any point such that $\text{ar}(\Delta AXB) =\text{ar}(\Delta XAC)$ , find the locus of $X$. I know that if sides $AB$, $AC$ and $BC$ are extended at a constant rate the position of $X$ remains the same. However I don't know what the locus of this would be .
(I haven't studied the topic locus I'm in class 9 any help would be highly appreciated.
 A: Let the origin of the axes coincide with $A$, let $B$ have coordinates $B(x_B,0)$ and $C$ have coordinates $C(x_C,y_C)$. I'll also call $P$ the point you called $X$, in order to avoid confusion. Let $P$ have coordinates $P(x,y)$. Considering distances from $P$ to $AC$ and $AB$, the requirement on the areas is readily shown to be equivalent to
$$|y x_B| = |y x_C - xy_C|,$$
that is
$$y^2x_B^2 = y^2x_C^2 -2xyx_Cy_C + x^2y_C^2,$$
leading to
\begin{equation}y^2(x_C^2-x_B^2) - 2xyx_Cy_C +x^2y_C^2 = 0.\tag{*}\label{eq}\end{equation}

If $x_B \neq \pm x_C$, solving with respect to $y$ yields
$$ y = \frac{xy_C(x_C \pm x_B)}{x_C^2-x_B^2},$$
leading to two straight lines with equations
$$r_1 : y = x \cdot\frac{y_C}{x_C-x_B},$$
and
$$r_2 : y = x \cdot\frac{y_C}{x_C+x_B}.$$

If $x_B = \pm x_C$ then the equation \eqref{eq} becomes
$$x(xy_C - 2yx_C) = 0,$$
so that the locus is given by the lines
$$r_1 : y = x \frac{y_C}{2x_C}$$
and
$$r_2 : x=0$$

In conclusion, the required locus is always given by two lines through $A$: the first one is the line containing the diagonal of the parallelogram determined by points $A$, $B$, and $C$; the second one is the line parallel to $BC$.

Below an example where $B(10,0)$ and $C(2,6)$, so that the two lines have equation
$$r_1 : y = \frac{1}{2}x$$
and
$$r_2: y =-\frac{3}{4}x$$


Finally an example where $x_C = x_B = 10$, $y_C = 6$ and locus is given therefore by the lines
$$r_1 : x = 0$$
and
$$r_2 : y = \frac{3}{10}x.$$

A: We start by drawing a triangle $\triangle ABC$, and we put a point $X$ somewhere inside. Draw perpendiculars from $X$ to the $AB$ and $AC$ sides, say to points $P$ and $Q$. They might be on the extensions. We write the area of $\triangle ABX$ as $\frac 12 AB\cdot XP$ and the area of $\triangle ACX$ as $\frac 12 AC\cdot XQ$. You get:
$$AB\cdot XP=AC\cdot XQ$$
We can write $$XP=AX\sin\angle XAP\\XQ=AX\sin\angle XAQ$$
By rearranging the terms,  you get: $$\frac{\sin\angle XAP}{\sin\angle XAQ}=\frac{AC}{AB}$$
Say $\angle XAQ=x$ and $\angle CAB=A$, then $\angle XAP=A-x$. Using $\sin(a-b)=\sin a\cos b-\cos a\sin b$, we get: $$\frac{\sin A\cos x-\cos A\sin x}{\sin x}=\frac{\sin A}{\tan x}-\cos A=\frac{AC}{AB}$$This yields $$\tan x=\frac{\sin A}{\cos A+\frac{AC}{AB}}$$
Since the right hand side is a constant, the locus of the points is a line that makes the angle $x$ with $AC$, where $x$ is given by the equation above.
