Question

The midpoint of the line joining the points A(8, 2) and B(4, 4) is _____​

Answer 1

Answer:

P(6,3)

Step-by-step explanation:

$\huge \boxed{\mathbb{ \color{violet} \colorbox{pink}{Given :-}}}$

$A(8,2) = x_1,y_1$

$B(4, 4) = x_2,y_2$

$\bold{Let \: the \: points \: be \: P}$

$\mathtt{Using \: the \: mid \: point \: formula,}$

$\frac{x2 + x1}{2} \: \: and \: \: \frac{y2 + y1}{2}$

$x = \frac{4 + 8}{2} , y = \frac{4 + 2}{2}$

$x = \frac{12}{2} ,y = \frac{6}{2}$

$\boxed{(x = 6,y = 3)}$

Hence the coordinates of point P are (6,3)

Answer 2

Given that, The Co-ordinates of endpoints of line are A(8, 2) and B(4, 4) .

Need To Find : Midpoint of the points A and B ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ $\sf Let's \:say \:that \:\pmb{\bf (x_m \:,\:y_m ) }\:be \:the \:midpoint \:of \: A \:and \: B \:$

Given that ,

• The Co-ordinates of endpoints of line are A(8, 2) and B(4, 4) .

Here ,

• x₁ = 8
• x₂ = 4
• y₁ = 2
• y₂ = 4

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ 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 , y₁ , y₂ are the points of line .

$\qquad:\implies \sf ( x_m \:,\: y_m) \:=\: \Bigg( \dfrac{(x_1 + x_2 )}{2} \:\:,\:\: \dfrac{(y_1 + y_2 )}{2}\:\:\Bigg) \:\\\\\qquad:\implies \sf ( x_m \:,\: y_m) \:=\: \Bigg(\dfrac{( 8 + 4 )}{2} \:\:,\:\: \dfrac{( 2 + 4 )}{2}\:\:\Bigg)\:\\\\\qquad:\implies \sf ( x_m \:,\: y_m) \:=\: \Bigg( \dfrac{( 12 )}{2} \:\:,\:\: \dfrac{( 6 )}{2}\:\:\Bigg) \:\\\\\qquad:\implies \sf ( x_m \:,\: y_m) \:=\: \Bigg(\dfrac{ 12 }{2} \:\:,\:\: \dfrac{ 6 }{2}\:\:\Bigg) \:\\\\ \qquad:\implies \underline {\boxed{\frak {( x_m \:,\: y_m) \:=\: ( 6 \:\:,\:\:3\:\:) \:}}}$

$\qquad \therefore \underline {\sf Hence, \:Midpoint \:of \:line \:A \:\:and \:B \:is \:\pmb{\bf ( \:6\:,\:3\:)\:}\:.}$