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प्रश्न
The centre of a circle is (a+2, a-1). Find the value of a, given that the circle passes through the points (2, -2) and (8, -2).
उत्तर
OA = OB [radii of same circle]
∴ OA2 = OB2
(a + 2 - 2)2 + (a - 1 + 2)2 = (a + 2 - 8)2 + (a - 1 + 2)2
a2 + (a + 1)2 = (a - 6)2 + (a + 1)2
a2 = a2 + 36 - 12a
12 a = 36
a = 3
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Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.
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Complete the table below the graph with the help of the following graph.
| Sr. No. | First point | Second point | Co-ordinates of first point (x1 , y1) | Co-ordinates of second point (x2 , y2) | `(y_2 - y_2)/(x_2 - x_2)` |
| 1 | C | E | (1, 0) | (3,4) | `4/2=2` |
| 2 | A | B | (-1,-4) | (0,-2) | `2/1 = 2` |
| 3 | B | D | (0,-2) | (2,2) | `4/2=2` |
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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`.
