Midpoints save the Day!

This is a simple English answer haha.

It is given that the coordinates are midpoints of the line. This is a major clue.

Imagine a 2nd triangle formed by these midpoints. That is, draw a line (a hypotenuse) with (3,5) as start and (-3,-3) as end. The two triangles are similar in size (you can do the proving for me lol).

As the coordinates are midpoints, then the lengths of the 2nd triangle are half the lengths of AC and AB. This means, the hypotenuse of the 2nd triangle is also half the length of BC.

Using the distance formula, you can calculate that the distance between the midpoints is 10 units. Therefore, BC is twice the length of 10 which gives the final answer of 20 units. BC=20

$\frac{x_A+x_B}{2}$ = $3$ ; by definition of midpoint. $\implies$ $x_B= 6-x_A$

$\frac{x_A+x_C}{2}$ = $-3$ ; by definition of midpoint $\implies$ $x_c=- 6-x_A$

$\frac{y_A+y_B}{2}$ = $5$ ; by definition of midpoint $\implies$ $y_B= 10-y_A$

$\frac{y_A+y_C}{2}$ = $-3$ ; by definition of midpoint $\implies$ $y_C= -6-y_A$

Thus: || $BC$ || = $\sqrt{(x_C-x_B)^{2}+(y_C-y_B)^{2})}$ = $\sqrt{(6+6)^{2}+(10+6)^{2})}$ = $20$

Let A,B,C be the vertices of the triangle and let the coordinates of A,B,C be (x1,y1), (x2, y2),(x3, y3) respectively. Applying midpiont theorm, Then (x1+x2)/2=3 which implies x1+x2=6--------(i)
(x1+x3)/2=-3 which implies x1+x3=-6--------(ii)
solving i and ii
x3-x2=12----------(I)

similarly,
(y1+y2)/2=5 which implies y1+y2=10------(i)
(y1+y3)/2=-3 which implies y1+y3=-6--------(ii)
solving (i) and (ii)
y3-y2=16--------------(II)

length of BC= root of (x3-x2)^2+(y3-y2)^2
which is 20 units