Advertisements
Advertisements
प्रश्न
Prove that the points (3, 0), (4, 5), (-1, 4) and (-2, -1), taken in order, form a rhombus.
Also, find its area.
Advertisements
उत्तर
The distance d between two points `(x_1, y_1)` and `x_2,y_2)` is given by the formula
`d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)`
In a rhombus, all the sides are equal in length. And the area ‘A’ of a rhombus is given as
`A = 1/2("Product of both diagonals")`
Here the four points are A(3,0), B(4,5), C(−1,4) and D(−2,−1).
First, let us check if all the four sides are equal.
`AB = sqrt((3 -4)^2 + (0 - 5)^2)`
`= sqrt((-1)^2 + (-5)^2)`
`= sqrt(1 + 25)`
`= sqrt(25 + 1)`
`BC=sqrt26`
`CD = sqrt((-1+2)^2 + (4 + 1)^2)`
`= sqrt((1)^2 +(5)^2)`
`= sqrt(26)`
`AD = sqrt((3 + 2)^2 + (0 + 1)^2)`
`= sqrt((5)^2 + (1)^2)`
`= sqrt(25 + 1)`
`AD = sqrt26`
Here, we see that all the sides are equal, so it has to be a rhombus.
Hence we have proved that the quadrilateral formed by the given four vertices is a rhombus.
Now let us find out the lengths of the diagonals of the rhombus.
`AC = sqrt((3 + 1)^2 + (0 - 4)^2)`
`= sqrt((4)^2 + (-4)^2)`
`= sqrt((6)^2 + (6)^2)`
`= sqrt(36 + 36)`
`BD = 6sqrt2`
Now using these values in the formula for the area of a rhombus we have,
`A = ((6sqrt2)(4sqrt2))/2`
`= ((6)(4)(2))/2`
A = 24
Thus the area of the given rhombus is 24 square units.
APPEARS IN
संबंधित प्रश्न
Find the points of trisection of the line segment joining the points:
(2, -2) and (-7, 4).
If A and B are (1, 4) and (5, 2) respectively, find the coordinates of P when AP/BP = 3/4.
Show that the points A (1, 0), B (5, 3), C (2, 7) and D (−2, 4) are the vertices of a parallelogram.
In what ratio does the point (−4, 6) divide the line segment joining the points A(−6, 10) and B(3,−8)?
If the points p (x , y) is point equidistant from the points A (5,1)and B ( -1,5) , Prove that 3x=2y
Show that the following points are the vertices of a rectangle.
A (2, -2), B(14,10), C(11,13) and D(-1,1)
Find the ratio in which the pint (-3, k) divide the join of A(-5, -4) and B(-2, 3),Also, find the value of k.
Show that the points (−2, 3), (8, 3) and (6, 7) are the vertices of a right triangle ?
If P (2, 6) is the mid-point of the line segment joining A(6, 5) and B(4, y), find y.
What is the distance between the points \[A\left( \sin\theta - \cos\theta, 0 \right)\] and \[B\left( 0, \sin\theta + \cos\theta \right)\] ?
If A (5, 3), B (11, −5) and P (12, y) are the vertices of a right triangle right angled at P, then y=
The distance of the point (4, 7) from the x-axis is
The coordinates of the fourth vertex of the rectangle formed by the points (0, 0), (2, 0), (0, 3) are
In which quadrant does the point (-4, -3) lie?
Find the point on the y-axis which is equidistant from the points (S, - 2) and (- 3, 2).
Write the equations of the x-axis and y-axis.
The points (–5, 2) and (2, –5) lie in the ______.
Point (3, 0) lies in the first quadrant.
In which ratio the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4)?
The distance of the point (–4, 3) from y-axis is ______.
