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Questions
Determine whether the points are collinear.
A(1, −3), B(2, −5), C(−4, 7)
Verify whether points A(1, −3), B(2, −5) and C(−4, 7) are collinear or not.
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Solution
Given: A(1, −3), B(2, −5), C(−4, 7)
Let,
A(1, −3) = A(x1, y1)
B(2, −5) = B(x2, y2)
C(−4, 7) = C(x3, y3)
By the distance formula,
d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`
= `sqrt((2 - 1)^2 + [-5 - (-3)]^2)`
= `sqrt((1)^2 + (-5 + 3)^2)`
= `sqrt((1)^2 + (-2)^2)`
= `sqrt(1+ 4)`
= `sqrt(5)` ...(1)
d(B, C) = `sqrt((x_3 - x_2)^2 + (y_3 - y_2)^2)`
= `sqrt((- 4 - 2)^2 + [7 - (-5)]^2)`
= `sqrt((-6)^2 + [7 + 5]^2)`
= `sqrt((-6)^2 + (12)^2)`
= `sqrt(36 + 144)`
= `sqrt(180)`
= `sqrt(36 xx 5)`
= `6sqrt(5)` ...(2)
d(A, C) = `sqrt((x_3 - x_1)^2 + (y_3 - y_1)^2)`
= `sqrt((-4 - 1)^2 + [7 - (-3)]^2)`
= `sqrt((-4 - 1)^2 + (7 + 3)^2)`
= `sqrt((-5)^2 + (10)^2)`
= `sqrt(25 + 100)`
= `sqrt(125)`
= `sqrt(25 × 5)`
= `5sqrt(5)` ...(3)
Adding (1) and (3)
∴ d(A, B) + d(A, C) = d(B, C)
∴ `sqrt5 + 5sqrt5 = 6sqrt5` ...(4)
∴ d(A, B) + d(A, C) = d(B, C) ...[From (2) and (4)]
∴ Points A(1, −3), B(2, −5) and C(−4, 7) are collinear.
Hence proved.
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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 y axis equidistant from B and C is ______.
A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6, 8) lies outside the circle.
The point P(–2, 4) lies on a circle of radius 6 and centre C(3, 5).
