Question

17)In ∆ ABC, the coordinates of Aare (3, 2) and the coordinates of the midpoint of AC and AB are (2, –1) and (1, 2) respectively. Find the coordinates of midpoint of BC.

Answer 1

Answer:

see the attachment.......

Answer 2

$Given \: in \: \triangle ABC , the$

$Coordinates \: of \: A(3,2) \:and \: the$

$mid \: point \: of \:AC \: is \: E(2,-1) \: and$

$mid \:point \: of \:AB \: are \: F(1,2) ,\: respectively.$

$Let \: Mid \:point \: of \: BC \: is \: D(x,y)$

$Coordinates \: of \: B(x_{1},y_{1}) \: and$ $C(x_{2},y_{2})$

$i) Mid \: point \:of\: AB = F(1,2)$

$\implies \Big(\frac{3+x_{1}}{2} , \frac{2+y_{1}}{2}\Big) = ( 1,2)$

$\implies \frac{3+x_{1}}{2} = 1 \:and \: \frac{2+y_{1}}{2} = 2$

$\implies 3 + x_{1} = 2 \:and \: 2+y_{1} = 4$

$\implies x_{1} = 2-3 \:and \: y_{1} = 4-2$

$\implies x_{1} = -1 \:and \: y_{1} = 2$

$\blue { \therefore B(x_{1} , y_{1}) = ( -1 , 2 ) }$

$ii) Mid \: point\:of \: AC = E(2,-1)$

$\implies \Big(\frac{3+x_{2}}{2} , \frac{2+y_{2}}{2}\Big) = ( 2,-1)$

$\implies \frac{3+x_{2}}{2} = 2 \:and \: \frac{2+y_{2}}{2} = -1$

$\implies 3 + x_{2} = 2 \:and \: 2+y_{2} = -1$

$\implies x_{2} = 4-3 \:and \: y_{2} = -2-2$

$\implies x_{2} = 1 \:and \: y_{1} = -4$

$\pink { \therefore C(x_{2} , y_{2}) = ( 1 , -4 ) }$

$iii)D(x,y) = Mid \: point\:of \: BC$

$= \Big(\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2}\Big)$

$= \Big( \frac{-1+1}{2} , \frac{2-4}{2}\Big)$

$= \Big( 0 , \frac{-2}{2}\Big)$

$= ( 0 , -1 )$

Therefore.,

$\red{ Coordinates \: of \: mid \:point \:BC }$

$\green { = ( 0 , -1 ) }$

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