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प्रश्न
The mid-point of the sides of a triangle are (2, 4), (−2, 3) and (5, 2). Find the coordinates of the vertices of the triangle
उत्तर
Let the vertices of the ΔABC be A(x1 y1), B(x2, y2) and C(x3, y3)
Mid−point of AB = `((x_1 + x_2)/2, (y_1 + y_2)/2)`
(2, 4) = `((x_1 + x_2)/2, (y_1 + y_2)/2)`
`(x_1 + x_2)/2` = 2
⇒ x1 + x2 = 4 → (1)
`(y_1 + y_2)/2` = 4
⇒ y1 + y2 = 8 → (2)
Mid−point of BC = `((x_2 + x_3)/2, (y_2 + y_3)/2)`
(−2, 3) = `((x_2 + x_3)/2, (y_2 + y_3)/2)`
`(x_2 + x_3)/2` = − 2
⇒ x2 + x3 = − 4 → (3)
`(y_2 + y_3)/2` = 3
⇒ y2 + y3 = 6 → (4)
Mid−point of AC = `((x_1 + x_3)/2, (y_1 + y_3)/2)`
(5, 2) = `((x_1 + x_3)/2, (y_1 + y_3)/2)`
`(x_1 + x_3)/2` = 5
⇒ x1 + x3 = 10 → (5)
`(y_1 + y_3)/2` = 2
⇒ y1 + y3 = 4 → (6)
By adding (1) + (3) + (5) we get
2x1 + 2x2 + 2x3 = 4 – 4 + 10
2(x1 + x2 + x3) = 10
⇒ x1 + x2 + x3 = 5
From (1) x1 + x2 = 4
⇒ 4 + x3 = 5
x3 = 5 – 4 = 1
∴ The vertices of the ΔABC are A(9, 3) B(– 5, 5), C(1, 1)
From (3) x2 + x3 = – 4
⇒ x1 + (– 4) = 5
x1 = 5 + 4 = 9
From (5) ⇒ x1 + x3 = 8
x2 + 10 = 5
x2 = 5 – 10 = – 5
∴ x1 = 9, x2 = – 5, x3 = 1
By adding (2) + (4) + (6) we get
2y1 + 2y2 + 2y3 = 8 + 6 + 4
2(y1 +y2 + y3) = 18
⇒ y1 + y2 + y3 = 9
From (2) ⇒ y1 + y2 = 8
8 + y3 = 9
⇒ y3 = 9 – 8
= 1
From (4)
y2 + y3 = 6
⇒ y1 + 6 = 9
y1 = 9 – 6
= 3
From (6)
y1 + y3 = 4
⇒ y2 + 4
= 9
y2 = 9 – 4
= 5
∴ y1 = 3, y2 = 5, y3 = 1
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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).
