Advertisements
Advertisements
प्रश्न
Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.
उत्तर
Given, AB = BC = CD
So, B is the mid-point of AC.
Let the co-ordinates of point A be (x, y).
∴ `(0, 3) = ((x + 1)/2, (y + 8)/2)`
`=> 0 = (x + 1)/2` and `3 = (y + 8)/2`
`=>` 0 = x + 1 and 6 = y + 8
`=>` –1 = x and –2 = y
Thus, the co-ordinates of point A are (–1, –2).
Also, C is the mid-point of BD.
Let the co-ordinates of point D be (p, q).
∴ `(1, 8) = ((0 + p)/2, (3 + q)/2)`
`=> 1 = (0 + p)/2` and `8 = (3 + q)/2`
`=>` 2 = 0 + p and 16 = 3 + q
`=>` 2 = p and 13 = q
Thus, the co-ordinates of point D are (2, 13).
APPEARS IN
संबंधित प्रश्न
(4, 2) and (-1, 5) are the adjacent vertices ofa parallelogram. (-3, 2) are the coordinates of the points of intersection of its diagonals. Find the coordinates of the other two vertices.
Find the centroid of a triangle whose vertices are (3, -5), (-7, 4) and ( 10, -2).
Two vertices of a triangle are (1, 4) and (3, 1). If the centroid of the triangle is the origin, find the third vertex.
P , Q and R are collinear points such that PQ = QR . IF the coordinates of P , Q and R are (-5 , x) , (y , 7) , (1 , -3) respectively, find the values of x and y.
The midpoint of the line segment joining the points P (2 , m) and Q (n , 4) is R (3 , 5) . Find the values of m and n.
Point M is the mid-point of segment AB. If AB = 8.6 cm, then find AM.
O(0, 0) is the centre of a circle whose one chord is AB, where the points A and B are (8, 6) and (10, 0) respectively. OD is the perpendicular from the centre to the chord AB. Find the coordinates of the mid-point of OD.
A(−3, 2), B(3, 2) and C(−3, −2) are the vertices of the right triangle, right angled at A. Show that the mid-point of the hypotenuse is equidistant from the vertices
Find coordinates of midpoint of the segment joining points (0, 2) and (12, 14)
Point P is the centre of the circle and AB is a diameter. Find the coordinates of points B if coordinates of point A and P are (2, – 3) and (– 2, 0) respectively.
Given: A`square` and P`square`. Let B (x, y)
The centre of the circle is the midpoint of the diameter.
∴ Mid point formula,
`square = (square + x)/square`
⇒ `square = square` + x
⇒ x = `square - square`
⇒ x = – 6
and `square = (square + y)/2`
⇒ `square` + y = 0
⇒ y = 3
Hence coordinates of B is (– 6, 3).
