Advertisements
Advertisements
प्रश्न
Prove that the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.
Advertisements
उत्तर
PQ = `sqrt((6 - 0)^2 + (2 + 4)^2) = 6sqrt(2)"units"`
QR = `sqrt((6 -3)^2 + (2 - 5)^2) = 3sqrt(2)"units"`
RS = `sqrt((3 +3)^2 + (5 + 1)^2) = 6sqrt(2)"units"`
PS = `sqrt((-3 - 0)^2 + (-1 + 4)^2) = 3sqrt(2)"units"`
PR = `sqrt((3 - 0)^2 + (5 + 4)^2) = 3sqrt(10)"units"`
QS = `sqrt((6 +3)^2 + (2 + 1)^2) = 3sqrt(10)"units"`
∵ PQ = RS and QR = PS,
Also PR = QS
∴ PQRS is a rectangle.
APPEARS IN
संबंधित प्रश्न
If the point P(2, 2) is equidistant from the points A(−2, k) and B(−2k, −3), find k. Also find the length of AP.
Find the distance between the points
(ii) A(7,-4)and B(-5,1)
Find the distance of the following points from the origin:
(iii) C (-4,-6)
Using the distance formula, show that the given points are collinear:
(6, 9), (0, 1) and (-6, -7)
Find the distances between the following point.
P(–6, –3), Q(–1, 9)
From the given number line, find d(A, B):

Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.
The length of line PQ is 10 units and the co-ordinates of P are (2, -3); calculate the co-ordinates of point Q, if its abscissa is 10.
By using the distance formula prove that each of the following sets of points are the vertices of a right angled triangle.
(i) (6, 2), (3, -1) and (- 2, 4)
(ii) (-2, 2), (8, -2) and (-4, -3).
If the length of the segment joining point L(x, 7) and point M(1, 15) is 10 cm, then the value of x is ______
