1. Set up coordinates

Take the square of side length $l$ with corners $A(0,0),\; B(l,0),\; C(l,l),\; D(0,l).$ Each corner has a mass $m$.

The diagonal $DB$ joins $D(0,l)$ and $B(l,0)$, so its equation is $x+y=l.$

An axis passing through $A(0,0)$ and parallel to $DB$ will therefore have equation $x+y=0,$ because it has the same slope as $DB$ and passes through the origin.

2. Use perpendicular distance formula

For a line $x+y=0,$ the perpendicular distance of a point $(x_1,y_1)$ from the line is $d=\frac{|x_1+y_1|}{\sqrt{1^2+1^2}}=\frac{|x_1+y_1|}{\sqrt 2}.$

Now find the distance of each mass from the axis.

• At $A(0,0)$: $d_A=\frac{|0+0|}{\sqrt 2}=0$

• At $B(l,0)$: $d_B=\frac{|l+0|}{\sqrt 2}=\frac{l}{\sqrt 2}$

• At $D(0,l)$: $d_D=\frac{|0+l|}{\sqrt 2}=\frac{l}{\sqrt 2}$

• At $C(l,l)$: $d_C=\frac{|l+l|}{\sqrt 2}=\frac{2l}{\sqrt 2}=\sqrt 2\,l$

3. Compute moment of inertia

Moment of inertia about the given axis is $I=\sum m d^2.$ So, $I=m(0)^2+m\left(\frac{l}{\sqrt 2}\right)^2+m\left(\frac{l}{\sqrt 2}\right)^2+m(\sqrt 2\,l)^2.$

Now simplify: $I=m\left(0+\frac{l^2}{2}+\frac{l^2}{2}+2l^2\right)$ $I=m(3l^2)$ $I=3ml^2.$

4. Match with options

The correct option is $\boxed{D\;:\; 3ml^2}$

5. Comparison with stored answer

Stored correct answer: $D$

Derived answer: $D$

So the derived answer agrees with the stored answer.