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Question
A line is of length 10 units and one end is at the point (2, – 3). If the abscissa of the other end be 10, prove that its ordinate must be 3 or – 9.
Solution
Let AB be the line of length 10 units then A (10, y)
A 10 units B
(2, -3) (10, y)
⇒ `sqrt((2 - 10)^2 + (-3 -y)^2)` = 100
Squaring both sides
⇒ 64 + 9 + y2 + 6y = 100
⇒ y2 + 6y + 73 - 100 = 0
⇒ y2 + 6y - 27 = 0
⇒ y2 + 9y - 3y - 27 = 0
⇒ y(y + 9) -3(y + 9) = 0
⇒ (y + 9)(y - 3) = 0
⇒ y = -9 or y = 3
Ordinate is 3 or -9.
Hence proved.
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