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प्रश्न
The points (2, –1), (–1, 4) and (–2, 2) are mid-points of the sides of a triangle. Find its vertices.
उत्तर
Let A(x1, y1), B(x2, y2) and C(x3, y3) be the coordinates of the vertices of ΔABC.
Midpoint of AB, i.e. D
`D(2, -1) = D((x_1 + x_2)/2, (y_1 + y_2)/2)`
`2 = (x_1 + x_2)/2, (y_1 + y_2)/2 = -1`
x1 + x2 = 4 ...(1)
y1 + y2 = –2 ...(2)
Similarly
x1 + x2 = –2 ...(3)
y1 + y3 = 8 ...(4)
x1 + x3 = −4 ...(5)
y2 + y3 = 4 ...(6)
Adding (1), (3) and (5), we get,
2(x1 + x2 + x3) = –2
x1 + x2 + x3 = –1
4 + x3 = –1 ...[From (1)]
x3 = –5
From (3)
x1 – 5 = –2
x1 = 3
From (5)
x2 – 5 = –4
x2 = 1
Adding (2), (4) and (6), we get,
2(y1 + y2 + y3) = 10
y1 + y2 + y3 = 5
–2 + y3 = 5 ...[From (2)]
y3 = 7
From (4)
y1 + 7 = 8
y1 = 1
From (6)
y2 + 7 = 4
y2 = –3
Thus, the co-ordinates of the vertices of ΔABC are (3, 1), (1, –3) and (–5, 7).
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| 1 | C | E | (1, 0) | (3,4) | `4/2=2` |
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