Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge Class 9

Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge Class 9

Question 1.

Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A(4, 7) and B(16, -2). Class 9

Solution:

A = (4, 7), B = (16, -2)

P and Q trisect AB

∴ AP = PQ = QB.

∴ P is the midpoint of AQ and Q is the midpoint of PB

Let P = (x₁ , y₁ ) and Q = (x₂, y₂)

P is midpoint of AQ

∴ x₁ = [latex]\frac{4+x_2}{2}[/latex] and y₁ = [latex]\frac{7+y_2}{2}[/latex]

∴ 2x₁ - x₂ = 4 and 2y₁ - y₂ = 7 ...(i)

Similarly Q is midpoint of PB

∴ x₂ = [latex]\frac{16+x_1}{2}[/latex] and y₂ = [latex]\frac{-2+y_1}{2}[/latex]

∴ x₁ = 2x₂ - 16 and y₁ = 2 + y₂ ...(ii)

Substituting (ii) in 1 (i), we get

2(2x₂ - 16) - x₂ = 4 and 2(2 + 2y₂) - y₂ = 7

∴ 3x₂ - 32 = 4 and 4 + 3y₂ = 7

∴ 3x₂ = 36 and 3y₂ = 3

∴ x₂ = 12 and y₂ = 1

From (i), we have

2x₁ - x₂ = 4 and 2y₁ - y₂ = 7

∴ 2x₁ - 12 = 4 and 2y₁ - 1 = 7

∴ 2x₁ = 16 and 2y₁ = 8

∴ x₁ = 8 and y₁ = 4

∴ P = (8, 4) and Q = (12, 1)