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प्रश्न
P( -2, 5), Q(3, 6 ), R( -4, 3) and S(-9, 2) are the vertices of a quadrilateral. Find the coordinates of the midpoints of the diagonals PR and QS. Give a special name to the quadrilateral.
उत्तर
Coordinates of mid point of PR are `((-2-4)/2 , (5 + 3)/2)` i.e. (-3 , 4)
Coordinates of mid point of QS are `((-9 + 3)/2 , (2 + 6)/2)` i.e. (-3 , 4)
The midpoint of PR is same as that of Qs , i.e. diagonals PR and QS bisect each other .
Hence , PQRS is a parallelogram.
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संबंधित प्रश्न
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Let A(-a, 0), B(0, a) and C(α , β) be the vertices of the L1 ABC and G be its centroid . Prove that
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A(3, 1), B(y, 4) and C(1, x) are vertices of a triangle ABC. P, Q and R are mid - points of sides BC, CA and AB respectively. Show that the centroid of ΔPQR is the same as the centroid ΔABC.
From the figure given alongside, find the length of the median AD of triangle ABC. Complete the activity.
Solution:
Here A(–1, 1), B(5, – 3), C(3, 5) and suppose D(x, y) are coordinates of point D.
Using midpoint formula,
x = `(5 + 3)/2`
∴ x = `square`
y = `(-3 + 5)/2`
∴ y = `square`
Using distance formula,
∴ AD = `sqrt((4 - square)^2 + (1 - 1)^2`
∴ AD = `sqrt((square)^2 + (0)^2`
∴ AD = `sqrt(square)`
∴ The length of median AD = `square`
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).
