Advertisements
Advertisements
प्रश्न
Using the distance formula, show that the given points are collinear:
(-1, -1), (2, 3) and (8, 11)
Advertisements
उत्तर
Let A(-1, -1) , B(2,3) and C( 8,11) be the give points. Then
`AB = sqrt((2+1)^2 +(3+1)^2) = sqrt((3)^2 +(4)^2) = sqrt(25)` =5 units
`BC=sqrt((8-2)^2 +(11-3)^2 = sqrt(6)^2 +(8)^2 = sqrt(100) `= 10 units
`AC = sqrt(( 8+1)^2 +(11+1)^2 ) = sqrt((9)^2 +(12)^2 = sqrt(225) `= 15 units
∴ AB +BC = (5+10) units = 15 units = Ac
Hence, the given points are collinear
APPEARS IN
संबंधित प्रश्न
If P (2, – 1), Q(3, 4), R(–2, 3) and S(–3, –2) be four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus
Find the coordinates of the centre of the circle passing through the points (0, 0), (–2, 1) and (–3, 2). Also, find its radius.
If Q (0, 1) is equidistant from P (5, − 3) and R (x, 6), find the values of x. Also find the distance QR and PR.
Find the distance of the following points from the origin:
(iii) C (-4,-6)
Using the distance formula, show that the given points are collinear:
(1, -1), (5, 2) and (9, 5)
If P (x , y ) is equidistant from the points A (7,1) and B (3,5) find the relation between x and y
Determine whether the points are collinear.
P(–2, 3), Q(1, 2), R(4, 1)
Find the distances between the following point.
R(–3a, a), S(a, –2a)
If A and B are the points (−6, 7) and (−1, −5) respectively, then the distance
2AB is equal to
Find the value of y for which the distance between the points A (3, −1) and B (11, y) is 10 units.
Find the distance between the following point :
(p+q,p-q) and (p-q, p-q)
Find the relation between a and b if the point P(a ,b) is equidistant from A (6,-1) and B (5 , 8).
Find the coordinate of O , the centre of a circle passing through A (8 , 12) , B (11 , 3), and C (0 , 14). Also , find its radius.
Prove that the points (1 , 1) , (-1 , -1) and (`- sqrt 3 , sqrt 3`) are the vertices of an equilateral triangle.
Find the distance between the following pair of points:
`(sqrt(3)+1,1)` and `(0, sqrt(3))`
The distance between the points (0, 5) and (–5, 0) is ______.
The point which lies on the perpendicular bisector of the line segment joining the points A(–2, –5) and B(2, 5) is ______.
Case Study -2
A hockey field is the playing surface for the game of hockey. Historically, the game was played on natural turf (grass) but nowadays it is predominantly played on an artificial turf.
It is rectangular in shape - 100 yards by 60 yards. Goals consist of two upright posts placed equidistant from the centre of the backline, joined at the top by a horizontal crossbar. The inner edges of the posts must be 3.66 metres (4 yards) apart, and the lower edge of the crossbar must be 2.14 metres (7 feet) above the ground.
Each team plays with 11 players on the field during the game including the goalie. Positions you might play include -
- Forward: As shown by players A, B, C and D.
- Midfielders: As shown by players E, F and G.
- Fullbacks: As shown by players H, I and J.
- Goalie: As shown by player K.
Using the picture of a hockey field below, answer the questions that follow:

The point on x axis equidistant from I and E is ______.
Point P(0, 2) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points A(–1, 1) and B(3, 3).
Find the value of a, if the distance between the points A(–3, –14) and B(a, –5) is 9 units.
