Find the distance of AB if point A is A(8,9) and the Midpoint MisM(-2, -3).

Answer: The distance of AB is [tex]4\sqrt{61}[/tex]Solution:In the question it is given that the midpoint M is (-2,-3) and point A is (8,9). To find the distance between A and B we have to find the point of B. Let the point of B be [tex](x_2,y_2)[/tex]We know,Mid point for [tex]x=\frac{x_{1}+x_{2}}{2}[/tex][tex]\Rightarrow-2=\frac{8+x_{2}}{2}[/tex][tex]\Rightarrow-4=8+x_{2}[/tex][tex]\Rightarrow-4-8=x_{2}[/tex][tex]\Rightarrow-12=x_{2}[/tex]Midpoint [tex]y=\frac{y_{2}+y_{2}}{2}[/tex][tex]\Rightarrow-3=\frac{9+y_{2}}{2}[/tex][tex]\Rightarrow-6=9+y_{2}[/tex][tex]\Rightarrow-6-9=y_{2}[/tex][tex]\Rightarrow-15=y_{2}[/tex]Therefore, the point of B is (-12,-15)Using the distance formula between two point A and B which is given by[tex]\mathrm{d}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}[/tex]where d is the distance[tex]d=\sqrt{(-12-8)^{2}+(-15-9)^{2}}[/tex][tex]d=\sqrt{(-20)^{2}+(-24)^{2}}[/tex][tex]d=\sqrt{400+576}[/tex][tex]d=\sqrt{976}[/tex][tex]\mathrm{d}=4 \sqrt{6} 1[/tex]