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Question
Point P is the centre of the circle and AB is a diameter. Find the coordinates of points B if coordinates of point A and P are (2, – 3) and (– 2, 0) respectively.
Given: A`square` and P`square`. Let B (x, y)
The centre of the circle is the midpoint of the diameter.
∴ Mid point formula,
`square = (square + x)/square`
⇒ `square = square` + x
⇒ x = `square - square`
⇒ x = – 6
and `square = (square + y)/2`
⇒ `square` + y = 0
⇒ y = 3
Hence coordinates of B is (– 6, 3).
Solution
Given: A(2, –3) and P(–2, 0). Let B (x, y)
The centre of the circle is the midpoint of the diameter.
∴ Mid point formula,
– 2 = `(bb2 + x)/bb2`
⇒ – 4 = 2 + x
⇒ x = – 4 – 2
⇒ x = – 6
and 0 = `(bb(-3) + y)/2`
⇒ – 3 + y = 0
⇒ y = 3
Hence coordinates of B is (– 6, 3).
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Solution:
Here A(–1, 1), B(5, – 3), C(3, 5) and suppose D(x, y) are coordinates of point D.
Using midpoint formula,
x = `(5 + 3)/2`
∴ x = `square`
y = `(-3 + 5)/2`
∴ y = `square`
Using distance formula,
∴ AD = `sqrt((4 - square)^2 + (1 - 1)^2`
∴ AD = `sqrt((square)^2 + (0)^2`
∴ AD = `sqrt(square)`
∴ The length of median AD = `square`
