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.
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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.
x1= 2, y1 = 5
x2 = -1, y2 = 2
m:n = 1:2
By section formula, x = (mx2+nx1)/(m+n)
x = (1×-1+2×2)/(1+2)
x = (-1+4)/(3)
x = 3/3 = 1
By section formula, y = (my2+ny1)/(m+n)
y = (1×2+2×5)/(1+2)
y = (2+10)/(3)
y = 12/3 = 4
Co-ordinates of P are (1,4).
Q(x,y) divides AC in the ratio 1:2.
x1= 2, y1 = 5
x2 = 5, y2 = 8
m:n = 1:2
By section formula, x = (mx2+nx1)/(m+n)
x = (1×5+2×2)/(1+2)
x = (5+4)/(3)
x = 9/3 = 3
By section formula, y = (my2+ny1)/(m+n)
y = (1×8+2×5)/(1+2)
y = (8+10)/(3)
y = 18/3 = 6
Co-ordinates of Q are (3,6).
(ii) By distance formula, d(PQ) = √[(x2-x1)2+(y2-y1)2]
Points are P(1,4) and Q(3,6).
So x1= 1, y1 = 4
x2 = 3, y2 = 6
d(PQ) = √[(x2-x1)2+(y2-y1)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) = √[(x2-x1)2+(y2-y1)2]
Points are B(-1,2) and C(5,8).
So x1= -1, y1 = 2
x2 = 5, y2 = 8
d(BC) = √[(x2-x1)2+(y2-y1)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.