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प्रश्न
(4, 2) and (-1, 5) are the adjacent vertices ofa parallelogram. (-3, 2) are the coordinates of the points of intersection of its diagonals. Find the coordinates of the other two vertices.
उत्तर
Let the coordinates of C and D be (x, y) and (a , b) respectively.
Midpoint of AC is O coordinates of O are ,
O (-3 , 2) = O `((4 + "x")/2 , (2 + "y")/2)`
`-3 = (4 + "x")/2 , 2 = (2 + "y")/2`
- 6 = 4 + x , 4 = 2 + y
x = - 10 , y = 2
C (-10 , 2)
Similarly , coordinates of midpoint of DB , i.e. O are ,
O (-3 , 2) = O `(("a" - 1)/2 , ("b" + 5)/2)`
`-3 = ("a" - 1)/2 , 2 = ("b" + 5)/2`
- 6 = a - 1 , 4 = b + 5
a = -5 , b = -1
D (- 5 , -1)
Thus , the coordinates of each other two vertices are (-10 , 2) and (-5 - 1)
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संबंधित प्रश्न
Given M is the mid-point of AB, find the co-ordinates of A; if M = (1, 7) and B = (–5, 10).
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Calculate the co-ordinates of the centroid of the triangle ABC, if A = (7, –2), B = (0, 1) and C =(–1, 4).
Find the coordinates of point P if P divides the line segment joining the points A(–1, 7) and B(4, –3) in the ratio 2 : 3.
In the following example find the co-ordinate of point A which divides segment PQ in the ratio a : b.
P(–2, –5), Q(4, 3), a : b = 3 : 4
Two vertices of a triangle are (1, 4) and (3, 1). If the centroid of the triangle is the origin, find the third vertex.
Find the mid-point of the line segment joining the points
`(1/2, (-3)/7)` and `(3/2, (-11)/7)`
A(−3, 2), B(3, 2) and C(−3, −2) are the vertices of the right triangle, right angled at A. Show that the mid-point of the hypotenuse is equidistant from the vertices
Point M (2, b) is the mid-point of the line segment joining points P (a, 7) and Q (6, 5). Find the values of ‘a’ and ‘b’.
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).
