Answer:
[tex] \huge{ \fbox{ \sf{( \: 3 \: , \: - 2)}}}[/tex]
Step-by-step explanation:
[tex] \star{ \: \sf{ \: Let \: the \: points \: be \: A \: and \: B}}[/tex]
[tex] \star{ \sf{ \: Let \: A(2, -3) \: be \: (x1 \:, y1) \: and \: B(4, -1) \: be \: (x2 ,\: y2)}}[/tex]
[tex] \underline{ \sf{Finding \: the \: midpoint}} : [/tex]
[tex] \boxed{ \sf{Midpoint = ( \frac{x1 + x2}{2} \: , \: \frac{y1 + y2}{2}) }}[/tex]
[tex] \mapsto{ \sf{Midpoint = ( \frac{2 + 4}{2} \: , \: \frac{ - 3 + ( - 1)}{2} }})[/tex]
[tex] \underline{ \text{Remember!}} : \sf{( + ) \times ( - ) = ( - )}[/tex]
[tex] \mapsto{ \sf{Midpoint = ( \frac{2 + 4}{2} \: , \: \frac{ - 3 - 1}{2} }})[/tex]
[tex] \underline{ \text{Remember!}} : \sf{The \: negative \: integers \: are \: always \: added \: but \: posses \: the \: negative( - ) \: sign.}[/tex]
[tex] \mapsto{ \sf{Midpoint = ( \frac{6}{2} \: , \: \frac{ - 4}{2} }})[/tex]
[tex] \underline{ \text{Remember!}} : \sf{Dividing \: a \: negative \: integer \: by \: a \: positive \: integer\: gives \: a \: negative \: integer.}[/tex]
[tex] \mapsto{ \boxed{ \sf{Midpoint = ( \: 3 \: , \: - 2})}}[/tex]
Hope I helped!
Best regards! :D
~TheAnimeGirl
