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A (10, 5), B (6, – 3) and C (2, 1) are the vertices of triangle ABC. L is the mid point of AB, M is the mid-point of AC. Write down the co-ordinates of L and M. Show that LM = 1/2 BC.

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ICSE Board Question Based on Section Formula Chapter of M.L Aggarwal for class10
In this question three vertices of a triangle are given.
Write down the coordinates asked in this question.
This is the Question Number 31, Exercise 11 of M.L Aggarwal.

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1. Given points are A(10,5), B(6,-3) and C(2,1).

Let L(x,y) be the midpoint of AB.

Here x1= 10, y1 = 5

x= 6, y2 = -3

By midpoint formula, x = (x1+x2)/2

x = (10+6)/2 = 16/2 = 8

By midpoint formula, y = (y1+y2)/2

y = (5-3)/2 = 2/2 = 1

So co-ordinates of L are (8,1).

Let M(x,y) be the midpoint of AC.

Here x1= 10, y1 = 5

x= 2, y2 = 1

By midpoint formula, x = (x1+x2)/2

x = (10+2)/2 = 12/2 = 6

By midpoint formula, y = (y1+y2)/2

y = (5+1)/2 = 6/2 = 3

So co-ordinates of M are (6,3).

By distance formula, d(LM) = √[(x2-x1)2+(y2-y1)2]

The points are L(8,1) and M(6,3)

So x1= 8, y1 = 1

x= 6, y2 = 3

d(LM) = √[(x2-x1)2+(y2-y1)2]

d(LM) = √[(6-8)2+(3-1)2]

d(LM) = √[(-2)2+(2)2]

d(LM) = √(4+4)

d(LM) = √8 = 2√2 …(i)

By distance formula, d(BC) = √[(x2-x1)2+(y2-y1)2]

The points are B(6,-3) and C(2,1).

So x1= 6, y1 = -3

x= 2, y2 = 1

d(BC) = √[(x2-x1)2+(y2-y1)2]

d(BC) = √[(2-6)2+(1-(-3))2]

d(BC) = √[(-4)2+(4)2]

d(BC) = √(16+16)

d(BC) = √32 = 4√2 …(ii)

From (i) and (ii), LM = ½ BC

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