Answer
$${\text{Point}}$$
Work Step by Step
$$\eqalign{
& 4{\left( {x - 3} \right)^2} + 3{\left( {y + 4} \right)^2} = 0 \cr
& {\text{Divide both sides by 12}} \cr
& \frac{{{{\left( {x - 3} \right)}^2}}}{3} + \frac{{{{\left( {y + 4} \right)}^2}}}{4} = 0 \cr
& {\text{The equation is written in the form }}\frac{{{{\left( {x - h} \right)}^2}}}{a} + \frac{{{{\left( {y - k} \right)}^2}}}{b} = {r^2} \cr
& {\text{Beacuse }}r = 0 \cr
& {\text{The graph of the equation }}4{\left( {x - 3} \right)^2} + 3{\left( {y + 4} \right)^2} = 0{\text{ is a Point}} \cr
& {\text{centered at }}\left( {3, - 4} \right) \cr} $$
