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Question
Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.
Solution
Given, AB = BC = CD
So, B is the mid-point of AC.
Let the co-ordinates of point A be (x, y).
∴ `(0, 3) = ((x + 1)/2, (y + 8)/2)`
`=> 0 = (x + 1)/2` and `3 = (y + 8)/2`
`=>` 0 = x + 1 and 6 = y + 8
`=>` –1 = x and –2 = y
Thus, the co-ordinates of point A are (–1, –2).
Also, C is the mid-point of BD.
Let the co-ordinates of point D be (p, q).
∴ `(1, 8) = ((0 + p)/2, (3 + q)/2)`
`=> 1 = (0 + p)/2` and `8 = (3 + q)/2`
`=>` 2 = 0 + p and 16 = 3 + q
`=>` 2 = p and 13 = q
Thus, the co-ordinates of point D are (2, 13).
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Find the coordinates of point P where P is the midpoint of a line segment AB with A(–4, 2) and B(6, 2).
Solution :
Suppose, (–4, 2) = (x1, y1) and (6, 2) = (x2, y2) and co-ordinates of P are (x, y).
∴ According to the midpoint theorem,
x = `(x_1 + x_2)/2 = (square + 6)/2 = square/2 = square`
y = `(y_1 + y_2)/2 = (2 + square)/2 = 4/2 = square`
∴ Co-ordinates of midpoint P are `square`.
