Advertisements
Advertisements
प्रश्न
A(6, -2), B(3, -2) and C(S, 6) are the three vertices of a parallelogram ABCD. Find the coordinates of the fourth vertex c.
उत्तर
We know that in a parallelogram, diagonals bisect each other .
∴ midpoint of AC = midpoint of BD
`"O" ((6 + 8)/2 , (-2 + 6)/2) = "O" (("x" + 3)/2 , ("y" - 2)/2)`
`therefore (6 + 8)/2 = ("x" + 3)/2 , (-2 + 6)/2 = ("y" - 2)/2`
14 = x + 3 , 4 = y - 2
x = 11 , y = 6
the coordinates of the fourth vertex Dare ( 11,6)
APPEARS IN
संबंधित प्रश्न
Given M is the mid-point of AB, find the co-ordinates of B; if A = (3, –1) and M = (–1, 3).
P(–3, 2) is the mid-point of line segment AB as shown in the given figure. Find the co-ordinates of points A and B.
Write the co-ordinates of the point of intersection of graphs of
equations x = 2 and y = -3.
If (-3, 2), (1, -2) and (5, 6) are the midpoints of the sides of a triangle, find the coordinates of the vertices of the triangle.
A triangle is formed by line segments joining the points (5, 1 ), (3, 4) and (1, 1). Find the coordinates of the centroid.
The coordinates of the end points of the diameter of a circle are (3, 1) and (7, 11). Find the coordinates of the centre of the circle.
The centre ‘O’ of a circle has the coordinates (4, 5) and one point on the circumference is (8, 10). Find the coordinates of the other end of the diameter of the circle through this point.
Find the coordinates of midpoint of segment joining (22, 20) and (0, 16)
From the figure given alongside, find the length of the median AD of triangle ABC. Complete the activity.
Solution:
Here A(–1, 1), B(5, – 3), C(3, 5) and suppose D(x, y) are coordinates of point D.
Using midpoint formula,
x = `(5 + 3)/2`
∴ x = `square`
y = `(-3 + 5)/2`
∴ y = `square`
Using distance formula,
∴ AD = `sqrt((4 - square)^2 + (1 - 1)^2`
∴ AD = `sqrt((square)^2 + (0)^2`
∴ AD = `sqrt(square)`
∴ The length of median AD = `square`
Find the coordinates of the mid-point of the line segment with points A(– 2, 4) and B(–6, –6) on both ends.
