smc

The right triangle has area $T$ and base $a$, so the height $h$ satisfies $\frac{1}{2}ha=T$. This means $h=\frac{2T}{a}$. Becuase the triangle is in the second quadrant, the coordinates are the origin, $(-a,0)$, and $(0,h)$, which means the third coordinate is $(0,\frac{2T}{a})$. So, we want the equation of a line though $(-a,0)$ and $(0,\frac{2T}{a})$. The slope is $\frac{\frac{2T}{a}-0}{0-(-a)}$, which simplifies to $\frac{2T}{a^2}$. The y-intercept is $(0,\frac{2T}{a})$, so the line in slope-intercept form is $y=mx+b$, or: $y=\frac{2T}{a^2}x+\frac{2T}{a}$ All the solutions have a positive x-coefficient and no fractions, so we clear the fractions by multiplying by $a^2$ and move the $y$ term to the right to get: $0=2Tx+2aT-a^{2}y$ This is equivalent to option $\boxed{2Tx-a^{2}y+2aT=0}$