Q.(2a, 0) and (0, a) are the extremities of the base of an isosceles triangle, and the equation of one of the equal sides is x = 2a. Find the equations of other two sides and the area of triangle.

The side $x=2a$ passes through $B=(2a,0)$, so the apex $A=(2a,y_A)$; using $AB=AC$ (isosceles) locates $A$, then the remaining side and base equations and the area follow.

Let base endpoints be $B=(2a,0)$ and $C=(0,a)$. Since the side $x=2a$ is vertical and passes through $B$ (as $B$ has $x$-coordinate $2a$), the apex $A$ lies on this line: $A=(2a,y_A)$, with $AB$ along $x=2a$.

Find $A$ using $AB=AC$ (isosceles condition):

$AB=|y_A|,\qquad AC=\sqrt{(2a)^2+(y_A-a)^2}.$

$y_A^2=4a^2+(y_A-a)^2=4a^2+y_A^2-2ay_A+a^2\implies0=5a^2-2ay_A\implies y_A=\dfrac{5a}{2}.$

So $A=\left(2a,\dfrac{5a}2\right)$.

Side $AC$ (from $A\left(2a,\frac{5a}2\right)$ to $C(0,a)$): slope $=\dfrac{5a/2-a}{2a-0}=\dfrac{3a/2}{2a}=\dfrac34$.

$y-a=\dfrac34(x-0)\implies3x-4y+4a=0.$

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