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प्रश्न
The midpoint of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a+1). Find the value of a and b.
show that the points A(- 1, 2), B(2, 5) and C(- 5, – 2) are collinear.
उत्तर
Midpoint of (2a , 4) and (-2 , 2b) is (1 , 2a + 1)
x = `(x_1 + x_2)/(2)`
1 = `(2a - 2)/(2)`
1 = a - 1
∴ a = 2
y = `(y_1 + y_2)/(2)`
2a + 1 = `(4 + 2b)/(2)`
2a + 1 = 2 + b
∴ 5 - 2 = b
∴ b = 3
Therefore, a = 2, b = 3.
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The points (2, –1), (–1, 4) and (–2, 2) are mid-points of the sides of a triangle. Find its vertices.
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The mid-point of the line segment joining A (- 2 , 0) and B (x , y) is P (6 , 3). Find the coordinates of B.
Two vertices of a triangle are ( -1, 4) and (5, 2). If the centroid is (0, 3), find the third vertex.
A lies on the x - axis amd B lies on the y -axis . The midpoint of the line segment AB is (4 , -3). Find the coordinates of A and B .
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)`
Find the co-ordinates of centroid of a triangle if points D(–7, 6), E(8, 5) and F(2, –2) are the mid-points of the sides of that triangle.
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).
