Answer: see proof below
Step-by-step explanation:
Since (a, b) is equidistant from (-a, 2) and (2, -b), then it is the midpoint of the those two points. Use Midpoint formula to find (a, b).
[tex]M_x=\dfrac{x_1+x_2}{2}\qquad \qquad \qquad M_y=\dfrac{y_1+y_2}{2}\\\\\\a=\dfrac{-a+2}{2}\qquad \qquad \qquad \quad b=\dfrac{2-b}{2}\\\\\\2a=-a+2\qquad \qquad \qquad \quad 2b=2-b\\\\\\3a=2\qquad \qquad \qquad \qquad \qquad 3b=2\\\\\\a=\dfrac{2}{3}\qquad \qquad \qquad \qquad \qquad b=\dfrac{2}{3}[/tex]
3(a + b) - 4 = 0
[tex]3\bigg(\dfrac{2}{3}+\dfrac{2}{3}\bigg)-4=0\\\\\\3\bigg(\dfrac{4}{3}\bigg)-4=0\\\\\\4-4=0\\\\0=0\qquad \text{TRUE!}[/tex]
Notice that I changed the equation to "negative 4" because the equation you provided did not make a true statement.
