Question

given the points (-4,8) and (6,-12) : determine the midpoint of the line segment connecting the points​

Answer 1

Given that, The Co-ordinates of endpoints of line are (-4 , 8 ) and  ( 6 , -12 ) .

Need To Find : Midpoint of the given endpoints of the line  ?

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❍ $\sf Let's \:say \:that \:\pmb{\bf (x_m \:,\:y_m ) }\:be \:the \:midpoint \:of \: line \:segment .\:$

Given that ,

⠀⠀⠀⠀▪︎ ⠀The Co-ordinates of endpoints of line are (-4, 8 ) and ( 6 , -12 ) .

$\qquad \dashrightarrow \sf \:( \: x_1 \:,\: y_1 \:) \:=\: ( \:-4\:,\:8\:)\:$

⠀⠀⠀⠀⠀&,

$\qquad \dashrightarrow \sf \:( \: x_2\:, \: y_2 \:) \:=\: ( \:6\:,\:-12\:)\:$

Here ,

• x₁ = -4

• x₂ = 6

• y₁ = 8

• y₂ = -12

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Formula for Midpoint is given by :

$\qquad \dag \:\:\:\:\underline {\boxed {\frak { \:\:\:Midpoint \:\:=\:\: \Bigg( \dfrac{(x_1 + x_2 )}{2} \:\:,\:\: \dfrac{(y_1 + y_2 )}{2}\:\:\Bigg)}}}$

⠀⠀⠀⠀⠀⠀Here ,⠀x₁ & x₂ are the Co-ordinates of x- axis  ,  y₁ & y₂ are the Co-ordinates of y - axis .

$\qquad:\implies \sf ( x_m \:,\: y_m) \:=\: \Bigg( \dfrac{(x_1 + x_2 )}{2} \:\:,\:\: \dfrac{(y_1 + y_2 )}{2}\:\:\Bigg) \:\\\\\qquad:\implies \sf ( x_m \:,\: y_m) \:=\: \Bigg(\dfrac{\{ (-4) + 6 \} }{2} \:\:,\:\: \dfrac{\{ 8 + (-12) \} }{2}\:\:\Bigg)\:\\\\\qquad:\implies \sf ( x_m \:,\: y_m) \:=\: \Bigg(\dfrac{\{ -4 + 6 \} }{2} \:\:,\:\: \dfrac{\{ 8 - 12 \} }{2}\:\:\Bigg)\:\\\\ \qquad:\implies \sf ( x_m \:,\: y_m) \:=\: \Bigg(\dfrac{\{ 2 \} }{2} \:\:,\:\: \dfrac{\{ -4 \} }{2}\:\:\Bigg)\:\\\\ \qquad:\implies \sf ( x_m \:,\: y_m) \:=\: \Bigg(\dfrac{ 2 }{2} \:\:,\:\: \dfrac{ -4 }{2}\:\:\Bigg)\:\\\\\qquad:\implies \underline {\boxed{\frak {( x_m \:,\: y_m) \:=\: ( 1 \:\:,\:\:-2\:\:) \:}}}$

$\qquad \therefore \underline {\sf Hence, \:Midpoint \:of \:line \: segment \:is \:\pmb{\bf ( \:1\:,\:-2\:)\:}\:.}$