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प्रश्न
Show that points A(–1, –1), B(0, 1), C(1, 3) are collinear.
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उत्तर १
A(–1, –1), B(0, 1), C(1, 3)
AB = `sqrt((0 + 1)^2 + (1 + 1)^2`
= `sqrt(1 + 4)`
AB = `sqrt(5)`
BC = `sqrt((1)^2 + (2)^2`
= `sqrt(1 + 4)`
BC = `sqrt(5)`
AC = `sqrt((2)^2 + (3 + 1)^2`
= `sqrt(4 + 16)`
= `sqrt(20)`
= `sqrt(5 xx 4)`
AC = `2sqrt(5)`
AB + BC = AC
`sqrt(5) + sqrt(5) = 2sqrt(5)`
A, B, and C are collinear.
उत्तर २
A`(x_1, y_1) = (-1, -1)`
B`(x_2, y_2) = (0,1)`
C`(x_3, y_3) = (1, 3)`
Slope of line
AB = `(y_2 - y_1)/(x_2 - x_1)`
= `(1 - (-1))/(0 - (-1))`
= `(1+1)/1 = 2`
Slope of line
BC = `(y_3 - y_2)/(x_3 - x_2)`
= `(3 - 1)/(1 - 0)`
= `2/1 = 2.`
As, slope of line AB = slope of line BC
Also AB and BC Hrtes contain common point B
∴ Points A, B, C are collinear. Hence Proved
APPEARS IN
संबंधित प्रश्न
If the point P(2, 2) is equidistant from the points A(–2, k) and B(–2k, –3), find k. Also, find the length of AP.
Find the distance between the following pairs of points:
(2, 3), (4, 1)
Name the type of quadrilateral formed, if any, by the following point, and give reasons for your answer:
(−3, 5), (3, 1), (0, 3), (−1, −4)
Two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of other two vertices.
A(–8, 0), B(0, 16) and C(0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5. Show that : PQ = `3/8` BC.
Find the distance between the points `A((-8)/5, 2)` and `B(2/5, 2)`.
Find x if distance between points L(x, 7) and M(1, 15) is 10.
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.
x (1,2),Y (3, -4) and z (5,-6) are the vertices of a triangle . Find the circumcentre and the circumradius of the triangle.
Find the distance between the following pairs of points:
`(3/5,2) and (-(1)/(5),1(2)/(5))`
A point P lies on the x-axis and another point Q lies on the y-axis.
Write the abscissa of point Q.
Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.
Point P (2, -7) is the center of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of: AT
Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.
Give the relation that must exist between x and y so that (x, y) is equidistant from (6, -1) and (2, 3).
Seg OA is the radius of a circle with centre O. The coordinates of point A is (0, 2) then decide whether the point B(1, 2) is on the circle?
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 ______.
Find the points on the x-axis which are at a distance of `2sqrt(5)` from the point (7, – 4). How many such points are there?
Find distance between points P(– 5, – 7) and Q(0, 3).
By distance formula,
PQ = `sqrt(square + (y_2 - y_1)^2`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(125)`
= `5sqrt(5)`
