Advertisements
Advertisements
प्रश्न
Show that the following points taken in order to form the vertices of a parallelogram
A(−7, −3), B(5, 10), C(15, 8) and D(3, −5)
Advertisements
उत्तर
AB = `sqrt((5 + 7)^2 + (10 + 3)^2`
= `sqrt((12)^2 + (13)^2`
= `sqrt(144 + 169)`
= `sqrt(313)`
BC = `sqrt((15 - 5)^2 + (8 - 10)^2`
= `sqrt(10^2 + (-2)^2`
= `sqrt(100 + 4)`
= `sqrt(104)`
CD = `sqrt((3 - 15)^2 + (-5 - 8)^2`
= `sqrt((-12)^2 + (- 13)^2`
= `sqrt(144 + 169)`
= `sqrt(313)`
AD = `sqrt((3 + 7)^2 + (-5 + 3)^2`
= `sqrt((10)^2 + (-2)^2`
= `sqrt(100 + 4)`
= `sqrt(104)`
AB = CD = `sqrt(313)` and BC = AD = `sqrt(104)` ...(Opposite sides are equal)
∴ ABCD is a parallelogram.
APPEARS IN
संबंधित प्रश्न
If the co-ordinate of A is x and that of B is y, find d(A, B).
x = 6, y = - 2
Find d(A, B), if co-ordinates of A and B are -2 and 5 respectively.
Co-ordinates of the pair of a point is given below. Hence find the distance between the pair.
3, 6
Co-ordinates of the pair of points are given below. Hence find the distance between the pair.
- 25, - 47
If P-Q-R and d(P, Q) = 3.4, d(Q, R)= 5.7 then d(P, R) = ?
Co-ordinate of point A on a number line is 1. What are the co-ordinates of points on the number line which are at a distance of 7 units from A?
Find the distance between the following pair of points
(3, −9) and (−2, 3)
Show that the following points taken in order to form an equilateral triangle
`"A"(sqrt(3), 2), "B"(0, 1), "C"(0, 3)`
If the points A(2, 0), B(– 6, 0), C(3, a – 3) lie on the x-axis then the value of a is _____
The distance between the point (5, −1) and the origin is _________
