Text Given A(−2,3)A(-2,3)A(−2,3) and B(4,−5)B(4,-5)B(4,−5), find the midpoint of ABABAB and the length ABABAB. Solution Midpoint: M(−2+42,3−52)=M(1,−1)M\left(\dfrac{-2+4}{2},\dfrac{3-5}{2}\right)=M(1,-1)M(2−2+4,23−5)=M(1,−1). Distance: AB=(4−(−2))2+(−5−3)2=36+64=100=10AB=\sqrt{(4-(-2))^{2}+(-5-3)^{2}}=\sqrt{36+64}=\sqrt{100}=10AB=(4−(−2))2+(−5−3)2=36+64=100=10. M(1,−1),AB=10 \boxed{\,M(1,-1),\quad AB=10\,}M(1,−1),AB=10