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 x_{1}= 10, y_{1}= 5x_{2 }= 6, y_{2}= -3By midpoint formula, x = (x

_{1}+x_{2})/2x = (10+6)/2 = 16/2 = 8By midpoint formula, y = (y

_{1}+y_{2})/2y = (5-3)/2 = 2/2 = 1So co-ordinates of L are (8,1).Let M(x,y) be the midpoint of AC.Here x_{1}= 10, y_{1}= 5x_{2 }= 2, y_{2}= 1By midpoint formula, x = (x

_{1}+x_{2})/2x = (10+2)/2 = 12/2 = 6By midpoint formula, y = (y

_{1}+y_{2})/2y = (5+1)/2 = 6/2 = 3So co-ordinates of M are (6,3).By distance formula, d(LM) = √[(x

_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]The points are L(8,1) and M(6,3)So x_{1}= 8, y_{1}= 1x_{2 }= 6, y_{2}= 3d(LM) = √[(x

_{2}-x_{1})^{2}+(y_{2}-y_{1})^{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) = √[(x

_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]The points are

B(6,-3) and C(2,1).So x_{1}= 6, y_{1}= -3x_{2 }= 2, y_{2}= 1d(BC) = √[(x

_{2}-x_{1})^{2}+(y_{2}-y_{1})^{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