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Maharashtra State BoardSSC (English Medium) 10th Standard

The Line Segment Ab is Divided into Five Congruent Parts at P, Q, R and S Such that A–P–Q–R–S–B. If Point Q(12, 14) and S(4, 18) Are Given Find the Coordinates of A, P, R, B. - Geometry Mathematics 2

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Question

The line segment AB is divided into five congruent parts at P, Q, R and S such that A–P–Q–R–S–B. If point Q(12, 14) and S(4, 18) are given find the coordinates of A, P, R, B.

Sum
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Solution

Points P, Q, R and S divide seg AB in five congruent parts.

Let A x1, y1), B(x2, y2), P(x3, y3) and R(x4, y4) be the given points.

Point R is the midpoint of segment QS. 

By midpoint formula,

x co-ordinate of R = `(12 + 4)/2 = 16/2` = 8

y co-ordinate of R = `(14 + 18)/2 = 32/2` = 16

∴ co-ordinates of R are (8, 16).
Point Q is the midpoint of seg PR.
By midpoint formula,
x co-ordinate of Q = `(x_3 + 8)/2`
∴ 12 = `(x_3 + 8)/2`
∴ 24 = (x3 + 8)
∴ x3 = 16
y co-ordinate of Q = `(y_3 + 16)/2`
∴ 14 = `(y_3 + 16)/2`
∴ 28 = y + 16
∴ y3 = 12
∴ P(x3, y3 ) = (16, 12)
∴ co-ordinates of P are (16, 12).
Point P is the midpoint of seg AQ.
By midpoint formula,
x co-ordinate of P = `(x_1 + 12)/2`
∴ 16 = `(x_1 + 12)/2`
∴ 32 = x1 + 12
∴  x1 = 20
y co-ordinate of P = `(y_1 + 14)/2`
∴ 12 = `(y_1 + 14)/2`
∴ 24 = y1 + 14
∴ y1 = 10
∴ A(x1, y1) = (20, 10)
∴ co-ordinates of A are (20, 10). Point S is the midpoint of seg RB.
By midpoint formula,
x co-ordinate of S = `(x_2 + 8)/2`
∴ 4 = `(x_2 + 8)/2`
∴ 8 = x2 + 8
∴  x2 = 0
y co-ordinate of S = `(y_2 + 16)/2`
∴ 18 = `(y_2 + 16)/2`
∴ 36 = y2 + 16
∴ y2 = 20
∴ B (x2, y2) = (0, 20)
∴ co-ordinates of B are (0, 20).
∴ The co-ordinates of points A, P, R and B are (20, 10), (16, 12), (8, 16) and (0, 20) respectively.
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Chapter 5: Co-ordinate Geometry - Problem Set 5 [Page 123]

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Balbharati Mathematics 2 [English] Standard 10 Maharashtra State Board
Chapter 5 Co-ordinate Geometry
Problem Set 5 | Q 19 | Page 123

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