An Important Question of class 10 Based on Section Formula Chapter of M.L Aggarwal for ICSE BOARD.

Here given three points are the vertices of a triangle ABC

Solve this question as asked.

This is the Question Number 32, Exercise 11 of M.L Aggarwal.

Deepak BoraNewbie

# A (2, 5), B ( – 1, 2) and C (5, 8) are the vertices of a triangle ABC. P and Q are points on AB and AC respectively such that AP : PB = AQ : QC = 1 : 2. (i) Find the co-ordinates of P and Q. (ii) Show that PQ = 1/3 BC

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(i) Given vertices of the ABC are

A(2,5), B(-1,2) and C(5,8).P and Q are points on AB and AC respectively such that AP:PB = AQ :QC = 1:2.P(x,y) divides AB in the ratio 1:2.x_{1}= 2, y_{1}= 5x_{2 }= -1, y_{2}= 2m:n = 1:2By section formula,x = (mx_{2}+nx_{1})/(m+n)x = (1×-1+2×2)/(1+2)x = (-1+4)/(3)x = 3/3 = 1By section formula,y = (my_{2}+ny_{1})/(m+n)y = (1×2+2×5)/(1+2)y = (2+10)/(3)y = 12/3 = 4Co-ordinates of P are (1,4).Q(x,y) divides AC in the ratio 1:2.x_{1}= 2, y_{1}= 5x_{2 }= 5, y_{2}= 8m:n = 1:2By section formula,x = (mx_{2}+nx_{1})/(m+n)x = (1×5+2×2)/(1+2)x = (5+4)/(3)x = 9/3 = 3By section formula,y = (my_{2}+ny_{1})/(m+n)y = (1×8+2×5)/(1+2)y = (8+10)/(3)y = 18/3 = 6Co-ordinates of Q are (3,6).(ii)By distance formula, d(PQ) = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]Points are P(1,4) and Q(3,6).So x_{1}= 1, y_{1}= 4x_{2 }= 3, y_{2}= 6d(PQ) = √[(x

_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]d(PQ) = √[(3-1)

^{2}+(6-4)^{2}]d(PQ) = √[(2)

^{2}+(2)^{2}]d(PQ) = √(4+4)

d(PQ) = √8 = 2√2 ..(i)

By distance formula, d(BC) = √[(x

_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]Points are B(-1,2) and C(5,8).So x_{1}= -1, y_{1}= 2x_{2 }= 5, y_{2}= 8d(BC) = √[(x

_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]d(BC) = √[(5-(-1))

^{2}+(8-2)^{2}]d(BC) = √[(6)

^{2}+(6)^{2}]d(BC) = √(36+36)

d(BC) = √72 = √(36×2) = 6√2 ..(ii)

BC/3 = 6√2/3 = 2√2 = PQ

PQ = 1/3 BC.Hence proved.