An airplane at T(80,20) needs to fly to both U(20,60) and V(110,85). What is the shortest possible distance for the trip? 165units 170units 97units 169 units is the answer A?

we know thatthe formula to calculate the distance between two points is equal to[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]Step [tex]1[/tex] Find the distance TU[tex]T(80,20)[/tex][tex]U(20,60)[/tex]substitute the values[tex]d=\sqrt{(60-20)^{2}+(20-80)^{2}}[/tex][tex]d=\sqrt{(40)^{2}+(-60)^{2}}[/tex][tex]d=\sqrt{1,600+3,600}[/tex][tex]d=\sqrt{5,200}\ units[/tex][tex]dTU=72\ units[/tex]Step [tex]2[/tex] Find the distanceUV[tex]U(20,60)[/tex][tex]V(110,85)[/tex]substitute the values[tex]d=\sqrt{(85-60)^{2}+(110-20)^{2}}[/tex][tex]d=\sqrt{(25)^{2}+(90)^{2}}[/tex][tex]d=\sqrt{625+8,100}[/tex][tex]d=\sqrt{8,725}\ units[/tex][tex]dUV=93\ units[/tex]Step [tex]3[/tex] Sum the distances TU and UV[tex]72\ units+93\ units=165\ units[/tex]thereforethe answer is the option 165 units