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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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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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
x2 = 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
x2 = 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
x2 = 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
x2 = 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