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प्रश्न
Points P(a, −4), Q(−2, b) and R(0, 2) are collinear. If Q lies between P and R, such that PR = 2QR, calculate the values of a and b.
उत्तर
Given, PR = 2QR
Now, Q lies between P and R, so, PR = PQ + QR
∴ PQ + QR = 2QR
`=>` PQ = QR
`=>` Q is the mid-point of PR.
∴ `(-2, b) = ((a + 0)/2, (-4 + 2)/2)`
`(-2, b) = (a/2, -1)`
`=>` a = −4 and b = –1
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संबंधित प्रश्न
Find the mid-point of the line segment joining the points:
(–6, 7) and (3, 5)
A(2, 5), B(1, 0), C(−4, 3) and D(–3, 8) are the vertices of quadrilateral ABCD. Find the co-ordinates of the mid-points of AC and BD. Give a special name to the quadrilateral.
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.
Point P is the midpoint of seg AB. If co-ordinates of A and B are (-4, 2) and (6, 2) respectively then find the co-ordinates of point P.
(A) (-1,2) (B) (1,2) (C) (1,-2) (D) (-1,-2)
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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.
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
The coordinates of diameter AB of a circle are A(2, 7) and B(4, 5), then find the coordinates of the centre
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 point P where P is the midpoint of a line segment AB with A(–4, 2) and B(6, 2).
Solution :
Suppose, (–4, 2) = (x1, y1) and (6, 2) = (x2, y2) and co-ordinates of P are (x, y).
∴ According to the midpoint theorem,
x = `(x_1 + x_2)/2 = (square + 6)/2 = square/2 = square`
y = `(y_1 + y_2)/2 = (2 + square)/2 = 4/2 = square`
∴ Co-ordinates of midpoint P are `square`.
