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प्रश्न
Show that A(1, 2), (1, 6), C(1 + 2`sqrt(3)`, 4) are vertices of an equilateral triangle.
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उत्तर
Distance between two points = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
By distance formula,
AB = `sqrt((1- 1)^2 + (6 - 2)^2`
= `sqrt(0^2 + 4^2)`
= `sqrt(4^2)`
= 4 ...(i)
BC = `sqrt((1 + 2sqrt(3) - 1)^2 + (4 - 6)^2`
= `sqrt((2sqrt(3))^2 + (-2)^2`
= `sqrt(12 + 4)`
= `sqrt(16)`
= 4 ...(ii)
AC = `sqrt((1 + 2sqrt(3) - 1)^2 + (4 -2)^2`
= `sqrt((2sqrt(3))^2 + 2^2`
= `sqrt(12 + 4)`
= `sqrt(16)`
= 4 ...(iii)
∴ AB = BC = AC ...[From (i), (ii) and (iii)]
∴ ∆ABC is an equilateral triangle.
∴ Points A, B and C are the vertices of an equilateral triangle.
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संबंधित प्रश्न
Prove that the points (–3, 0), (1, –3) and (4, 1) are the vertices of an isosceles right angled triangle. Find the area of this triangle
Find the distance between the following pairs of points:
(−5, 7), (−1, 3)
ABC is a triangle and G(4, 3) is the centroid of the triangle. If A = (1, 3), B = (4, b) and C = (a, 1), find ‘a’ and ‘b’. Find the length of side BC.
Find the value of a when the distance between the points (3, a) and (4, 1) is `sqrt10`
Find the distance between the points:
A(7, –4) and B(–5, 1)
Find all possible values of y for which the distance between the points A(2, –3) and B(10, y) is 10 units.
Find the distances between the following point.
P(–6, –3), Q(–1, 9)
If A and B are the points (−6, 7) and (−1, −5) respectively, then the distance
2AB is equal to
Prove that the following set of point is collinear :
(4, -5),(1 , 1),(-2 , 7)
Prove that the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.
Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.
KM is a straight line of 13 units If K has the coordinate (2, 5) and M has the coordinates (x, – 7) find the possible value of x.
Find distance between point Q(3, –7) and point R(3, 3)
Solution: Suppose Q(x1, y1) and point R(x2, y2)
x1 = 3, y1 = –7 and x2 = 3, y2 = 3
Using distance formula,
d(Q, R) = `sqrt(square)`
∴ d(Q, R) = `sqrt(square - 100)`
∴ d(Q, R) = `sqrt(square)`
∴ d(Q, R) = `square`
Find distance between point A(–1, 1) and point B(5, –7):
Solution: Suppose A(x1, y1) and B(x2, y2)
x1 = –1, y1 = 1 and x2 = 5, y2 = –7
Using distance formula,
d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
∴ d(A, B) = `sqrt(square +[(-7) + square]^2`
∴ d(A, B) = `sqrt(square)`
∴ d(A, B) = `square`
The distance between the points A(0, 6) and B(0, -2) is ______.
A circle drawn with origin as the centre passes through `(13/2, 0)`. The point which does not lie in the interior of the circle is ______.
If the distance between the points (4, P) and (1, 0) is 5, then the value of p 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 ______.
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In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.
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Based on the above information answer the following questions using the coordinate geometry.
- Find the distance between Lucknow (L) to Bhuj (B).
- If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
- Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P)
[OR]
Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).
Show that points A(–1, –1), B(0, 1), C(1, 3) are collinear.
