Question

A (5,3), B(-1,1) and C(7,-3)are the vertices of triangle ABC. If l is the mid-point of AB and M is the mid point of AC, show that LM=1/2BC
urgent plzzz

Answer 1

L = [ (5-1)/2, (3+1)/2 ] = (2, 2)                   formula for midpoint: [ (x1+x2)/2 ,  (y1+y2)/2 ]
M = [ (5+7)/2, (3-3)/2 ] = (6, 0)

LM² = (6-2)² + (0-2)² = 20             =>   LM = 2√5     as    distance² = (x1-x2)²+(y1-y2)²

BC² = (7+1)² + (-3-1)² = 80    =>   BC = 4√5

Hence  LM = BC/2

Answer 2

Answer:

LM = 2 √5

BC = 4 √5

Step-by-step explanation :-

Mid - point Formula -

(x₁ + x₂ / 2 , y₁ + y₂ / 2)

L = ( 5 - 1 /2 , 3 + 1 / 2 )

  = (4 / 2 , 4 / 2)

  = (2,2)

M= (5 + 7 / 2 , 3 - 3  / 2 )

   =  (12 / 2 , 0 / 2 )

   = (6,0)

Distance formula :-

( x₁ + x₂ )² + ( y₁ + y₂)²

LM² = ( x₁ + x₂ )² + ( y₁ + y₂)²

       = ( 6-2)² + (0-2)²

       = ( 4 )² + ( -2 )²

       = 16 + 4

       =  20

BC² = ( x₁ + x₂ )² + ( y₁ + y₂)²

       = ( 7+ 1 )² + (-3 - 1 )²

       = (8)² + ( -4 )²

       = 64 + 16

       = 80

Therefore ,

BC = √80

     = 4√5

Hence ,

LM = 1 / 2 BC