In an examination hall, students are seated at a distance from each other, to maintain the social distance due to CORONA virus pandemic. Let three students sit at points 𝐀, 𝐁 and 𝐂 whose coordinates are (𝟐, −𝟑), (𝟒,𝟐), and (𝟖, 𝟓) respectively. Based on the above information, answer the following questions.
1. What is the distance between 𝐀 and 𝐂?
2. What is the midpoint of line segment joining 𝐀 and 𝐂?
3. If an invigilator at the point 𝐈, lying on the straight line joining 𝐁 and 𝐂 such that it divides the distance between them in the ratio of 𝟐 :𝟑. Then what are the coordinates of 𝐈?

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Solution

Soluton 1.
The distance between A and C:
AC = √[(8 - 2)^2 + (5 + 3)^2]
AC = √(36 +64) = √100
AC = 10 unit
Soluton 2.
The midpoint of A and C can be calculated by the midpoint formula as
(2 + 8)/2, (-3 + 5)/2 = (5, 1)
Hence midpoint of AC is (5, 1).
Soluton 3.
Let the coordinates of I be (x, y).
Then, by section formula
Dictate and write:
x = [(2×8) + (3×4)]/(2 +3)
x = [16 + 12]/5 = 28/5
And
y = [(2×5) + (3×2)]/(2+3)
y = [10 + 6]/5 = 16/5
Hence, the coordinate of I are (28/5, 16/5).

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Q. In an examination hall, students are seated at a distance from each other, to maintain the social distance due to CORONA virus pandemic. Let three students sit at points 𝐀, 𝐁 and 𝐂 whose coordinates are (𝟐, −𝟑), (𝟒,𝟐), and (𝟖, 𝟓) respectively.

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