Q.Find the equation of the line with slope $\dfrac{1}{2}$ passing through $(-4, 3)$.

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 $=\dfrac{5a/2-a}{2a-0}=\dfrac{3a/2}{2a}=\dfrac34$.

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

Base $BC$ (from $B(2a,0)$ to $C(0,a)$): slope $=\dfrac{a-0}{0-2a}=-\dfrac12$.

$y=-\dfrac12(x-2a)\implies x+2y-2a=0.$

Area (using vertices $A\left(2a,\frac{5a}2\right),B(2a,0),C(0,a)$):

$\text{Area}=\dfrac12\left|2a(0-a)+2a\left(a-\dfrac{5a}2\right)+0\left(\dfrac{5a}2-0\right)\right|=\dfrac12\left|-2a^2-3a^2\right|=\dfrac12(5a^2)=\dfrac{5a^2}{2}.$