Advertisements
Advertisements
Question
Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.
Advertisements
Solution
Let the given points be A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4).
AB = `sqrt((-5 + 3)^2 + (-5 - 2)^2) = sqrt(4+49) = sqrt(53)`
BC = = `sqrt((2+5)^2 + (-3+5)^2) = sqrt(49+4) = sqrt(53)`
CD= = `sqrt((4 - 2)^2 + (4+3)^2) = sqrt(4 + 49) = sqrt(53)`
DA = `sqrt((-3 - 4)^2 + (2 - 4)^2) = sqrt(49+4) = sqrt(53)`
AC =`sqrt((2+3)^2 + (-3 - 2)^2) = sqrt(25+25) = 5sqrt(2)`
BD =`sqrt((4+5)^2 + (4+5)^2) = sqrt(81+ 81) = 9sqrt(2)`
Since, AB = BC = CD = DA and AC ≠ BD
The given vertices are the vertices of a rhombus.
APPEARS IN
RELATED QUESTIONS
Find the distance between the following pair of points:
(-6, 7) and (-1, -5)
`" Find the distance between the points" A ((-8)/5,2) and B (2/5,2)`
A line segment of length 10 units has one end at A (-4 , 3). If the ordinate of te othyer end B is 9 , find the abscissa of this end.
Points A (-3, -2), B (-6, a), C (-3, -4) and D (0, -1) are the vertices of quadrilateral ABCD; find a if 'a' is negative and AB = CD.
Calculate the distance between the points P (2, 2) and Q (5, 4) correct to three significant figures.
Find the distance of the following points from origin.
(a cos θ, a sin θ).
Use distance formula to show that the points A(-1, 2), B(2, 5) and C(-5, -2) are collinear.
Give the relation that must exist between x and y so that (x, y) is equidistant from (6, -1) and (2, 3).
Find distance of point A(6, 8) from origin.
A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6, 8) lies outside the circle.
