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प्रश्न
Two vertices of a triangle are (1, 4) and (3, 1). If the centroid of the triangle is the origin, find the third vertex.
उत्तर
Given the centroid of Δ ABC is at origin , i.e. G (0 , 0).
Let the coordinates of third vertex be (x , y).
Coordinates of G are ,
G (0 , 0) = G `((1 + 3 + "x")/3 , (4 + 1 + "y")/3)`
O = `(4 + "x")/2` , O = `(5 + "y")/2`
x = - 4 , y = -5
Coordinates of third vertex are (-4 , -5)
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संबंधित प्रश्न
Find the mid-point of the line segment joining the points:
(–6, 7) and (3, 5)
Prove that the points A(–5, 4); B(–1, –2) and C(5, 2) are the vertices of an isosceles right-angled triangle. Find the co-ordinates of D so that ABCD is a square.
Find the coordinates of point P if P divides the line segment joining the points A(–1, 7) and B(4, –3) in the ratio 2 : 3.
Find th co-ordinates of the midpoint of the line segment joining P(0, 6) and Q(12, 20).
A( 4, 2), B(-2, -6) and C(l, 1) are the vertices of triangle ABC. Find its centroid and the length of the median through C.
The coordinates of the centroid I of triangle PQR are (2, 5). If Q = (-6, 5) and R = (7, 8). Calculate the coordinates of vertex P.
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.
Find the mid-point of the line segment joining the points
`(1/2, (-3)/7)` and `(3/2, (-11)/7)`
Find coordinates of the midpoint of a segment joining 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 midpoint formula,
∴ Coordinates of midpoint of segment AB
= `((x_1 + x_2)/2, (y_1+ y_2)/2)`
= `(square/2, square/2)`
∴ Coordinates of the midpoint = `(4/2, square/2)`
∴ Coordinates of the midpoint = `(2, square)`
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).
