Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
AB = CD
AB2 = CD2
(- 6 + 3)2 + (a + 2)2 = (0 + 3)2 + (- 1 + 4)2
9 + a2 + 4 + 4a = 9 + 9
a2 + 4a - 5 = 0
a2 - a + 5a - 5 = 0
a(a - 1) + 5 (a - 1) = 0
(a - 1) (a + 5) = 0
a = 1 or - 5
It is given that a is negative, thus the value of a is - 5.
APPEARS IN
संबंधित प्रश्न
Show that four points (0, – 1), (6, 7), (–2, 3) and (8, 3) are the vertices of a rectangle. Also, find its area
Find the distance between the points:
A(7, –4) and B(–5, 1)
Find all possible values of x for which the distance between the points A(x, –1) and B(5, 3) is 5 units.
Using the distance formula, show that the given points are collinear:
(–1, –1), (2, 3) and (8, 11)
Find the distance between the following pairs of point in the coordinate plane :
(4 , 1) and (-4 , 5)
Find the coordinates of O, the centre passing through A( -2, -3), B(-1, 0) and C(7, 6). Also, find its radius.
Given A = (3, 1) and B = (0, y - 1). Find y if AB = 5.
Calculate the distance between A (7, 3) and B on the x-axis, whose abscissa is 11.
Show that the point (0, 9) is equidistant from the points (–4, 1) and (4, 1).
Find the value of a, if the distance between the points A(–3, –14) and B(a, –5) is 9 units.
