​​先阅读一段文字,再回答下列问题:已知在平面内两点坐标$P_{1}(x_{1}$,$y_{1})$,$P_{2}(x_{2}$,$y_{2})$,其两点间距离公式为:$p_{1}p_{2}=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$,例如:点$\left(3,2\right)$和$\left(4,0\right)$的距离为$\sqrt{(3-4)^{2}+(2-0)^{2}}=\sqrt{5}$.同时,当两点所在的直线在坐标轴上或平行于$x$轴或平行于$y$轴距离公式可简化成:$p_{1}p_{2}=|x_{1}-x_{2}|$或$p_{1}p_{2}=|y_{1}-y_{2}|$.
$(1)$已知$A$、$B$在平行于$y$轴的直线上,点$A$的纵坐标为$5$,点$B$的纵坐标为$2$,则$A$,$B$两点的距离为______；
$(2)$线段$AB$平行于$x$轴,且$AB=3$,若点$B$的坐标为$\left(2,4\right)$,则点$A$的坐标是______；
$(3)$已知$A\left(3,5\right)$,$B\left(-4,4\right)$,$A$,$B$两点的距离为______；
$(4)$已知$\triangle ABC$三个顶点坐标为$A\left(3,4\right)$,$B\left(0,5\right)$,$C\left(-1,2\right)$,请判断此三角形的形状,并说明理由.