Advertisements
Advertisements
प्रश्न
Show that the points A (3,1) , B (0,-2) , C(1,1) and D (4,4) are the vertices of parallelogram ABCD.
उत्तर
The points are A (3,1) , B (0,-2) , C(1,1) and D (4,4)
Join AC and BD, intersecting at O.
We know that the diagonals of a parallelogram bisect each other `".Midpoint of AC" = ((3+1)/2 , (1+1)/2) = (4/2,2/2) = (2,1) `
`"Midpoint of BD " ((0+4)/2 , (-2+4)/4) = (4/2,2/2) = (2,1)`
Thus, the diagonals AC and BD have the same midpoint
Therefore, ABCD is a parallelogram.
APPEARS IN
संबंधित प्रश्न
If the points A(−1, −4), B(b, c) and C(5, −1) are collinear and 2b + c = 4, find the values of b and c.
If D, E and F are the mid-points of sides BC, CA and AB respectively of a ∆ABC, then using coordinate geometry prove that Area of ∆DEF = `\frac { 1 }{ 4 } "(Area of ∆ABC)"`
Find the centre of a circle passing through the points (6, − 6), (3, − 7) and (3, 3).
Find the area of the following triangle:
Prove that the points (a, 0), (0, b) and (1, 1) are collinear if `1/a+1/b=1`
Find the area of a triangle whose sides are 9 cm, 12 cm and 15 cm ?
Show that the points (-3, -3),(3,3) and C (-3 `sqrt(3) , 3 sqrt(3))` are the vertices of an equilateral triangle.
Find the value of k so that the area of the triangle with vertices A (k+1, 1), B(4, -3) and C(7, -k) is 6 square units
The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is ______.
Find the missing value:
Base | Height | Area of parallelogram |
______ | 15 cm | 154.5 cm2 |