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Question
Show that the points (2, 0), (–2, 0) and (0, 2) are vertices of a triangle. State the type of triangle with reason.
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Solution
Let the points be P(2, 0), Q(–2, 0) and R(0, 2).
Distance between two points = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
By distance formula,
d(P, Q) = `sqrt([(-2) - 2]^2 + (0 - 0)^2`
= `sqrt((-4)^2 + (0)^2`
= `sqrt(16 + 0)`
= 4 ...(i)
d(Q, R) = `sqrt([0 - (-2)]^2 + (2 - 0)^2`
= `sqrt((2)^2 + (2)^2`
= `sqrt(4 + 4)`
= `sqrt(8)` ...(ii)
d(P, R) = `sqrt((0 -2)^2 + (2 - 0)^2`
= `sqrt((- 2)^2 + (2)^2`
= `sqrt(4 + 4)`
= `sqrt(8)` ...(iii)
On adding (ii) and (iii),
d(P, Q) + d(Q, R) = `4 + sqrt(8)`
`4 + sqrt(8) > sqrt(8)`
∴ d(P, Q) + d(Q, R) > d(P, R)
∴ Points P, Q, R are non colinear points.
We can construct a triangle through 3 non collinear points.
∴ The segment joining the given points form a triangle.
Since P(Q, R) = P(P, R)
∴ ∆PQR is an isosceles triangle.
∴ The segment joining the points (2, 0), (–2, 0) and (0, 2) will form an isosceles triangle.
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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 ______.
If (– 4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.
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The point P(–2, 4) lies on a circle of radius 6 and centre C(3, 5).
The points A(–1, –2), B(4, 3), C(2, 5) and D(–3, 0) in that order form a rectangle.
