Advertisements
Advertisements
Question
Prove that the points (4 , 6) , (- 1 , 5) , (- 2, 0) and (3 , 1) are the vertices of a rhombus.
Solution
AB = `sqrt ((4 + 1)^2 + (6 - 5)^2) = sqrt (25 + 1) = sqrt 26` units
BC = `sqrt ((-1 + 2)^2 + (5 - 0)^2) = sqrt (1 + 25) = sqrt 26` units
CD = `sqrt ((-2 - 3)^2 + (0 - 1)^2) = sqrt (25 + 1) = sqrt 26` units
DA = `sqrt ((3 - 4)^2 + (1 - 6)^2) = sqrt (1 + 25) = sqrt 26` units
AC = `sqrt ((4 + 2)^2 + (6 - 0)^2) = sqrt (36 + 36) = 36 sqrt 2` units
BD = `sqrt ((-1-3)^2 + (5 - 1)^2) = sqrt (36 + 36) = 16 sqrt 2` units
∵ AB = BC = CD = DA and AC ≠ BD
∴ ABCD is a rhombus.
APPEARS IN
RELATED QUESTIONS
Find the values of y for which the distance between the points P (2, -3) and Q (10, y) is 10 units.
Find the distance of a point P(x, y) from the origin.
Given a triangle ABC in which A = (4, −4), B = (0, 5) and C = (5, 10). A point P lies on BC such that BP : PC = 3 : 2. Find the length of line segment AP.
Find the value of y for which the distance between the points A (3, −1) and B (11, y) is 10 units.
The distance between the points (3, 1) and (0, x) is 5. Find x.
Given A = (3, 1) and B = (0, y - 1). Find y if AB = 5.
Find distance of point A(6, 8) from origin
Find distance between point A(–1, 1) and point B(5, –7):
Solution: Suppose A(x1, y1) and B(x2, y2)
x1 = –1, y1 = 1 and x2 = 5, y2 = – 7
Using distance formula,
d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
∴ d(A, B) = `sqrt(square +[(-7) + square]^2`
∴ d(A, B) = `sqrt(square)`
∴ d(A, B) = `square`
The point A(2, 7) lies on the perpendicular bisector of line segment joining the points P(6, 5) and Q(0, – 4).
A point (x, y) is at a distance of 5 units from the origin. How many such points lie in the third quadrant?
