Question

Find the distance between P(7,2) and Q(2,5) to the nearest tenth.

Answer 1

`P(7,2) \\ x_1=7 \\ y_1=2 \\ \\ Q(2,5) \\ x_2=2 \\ y_2=5 \\ \\ \hbox{the distance between the points:} \\ d=√((x_2-x_1)^2+(y_2-y_1)^2)=√((2-7)^2+(5-2)^2)= \\ √((-5)^2+3^2)=√(25+9)=√(34) \approx 5.83`

The distance is 5.83 units.