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प्रश्न
ABC is a triangle whose vertices are A(1, –1), B(0, 4) and C(– 6, 4). D is the midpoint of BC. Find the coordinates of D.
उत्तर
D is the mid-point of BC.
∴ Co-ordinates of D are `((0 - 6)/2, (4 + 4)/2)`
i.e., (–3, 4)
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संबंधित प्रश्न
Find the mid-point of the line segment joining the points:
(5, –3) and (–1, 7)
Calculate the co-ordinates of the centroid of the triangle ABC, if A = (7, –2), B = (0, 1) and C =(–1, 4).
The co-ordinates of the centroid of a triangle PQR are (2, –5). If Q = (–6, 5) and R = (11, 8); calculate the co-ordinates of vertex P.
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The three vertices of a parallelogram taken in order are (-1, 0), (3, 1) and (2, 2) respectively. Find the coordinates of the fourth vertex.
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
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`
Point M (2, b) is the mid-point of the line segment joining points P (a, 7) and Q (6, 5). Find the values of ‘a’ and ‘b’.
