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प्रश्न
Show that the ▢PQRS formed by P(2, 1), Q(–1, 3), R(–5, –3) and S(–2, –5) is a rectangle.
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उत्तर
Given: P(2, 1), Q(–1, 3), R(–5, –3) and S(–2, –5)
Distance Formula = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`
PQ = `sqrt((-1 - 2)^2 + (3 - 1)^2)`
= `sqrt((3-)^2 + (2)^2)`
= `sqrt(9 + 4)`
= `sqrt 13` ...(i)
QR = `sqrt([-5 - (-1)]^2 + (-3 - 3)^2)`
= `sqrt((-4)^2 + (-6)^2)`
= `sqrt(16 + 36)`
= `sqrt 52`
= `sqrt(2 xx 2 xx 13)`
= 2`sqrt13` ...(ii)
RS = `sqrt([-2 - (-5)]^2 + [-5 - (-3)]^2)`
= `sqrt((-2 + 5)^2 + (-5 + 3)^2)`
= `sqrt(3^2 + (-2)^2)`
= `sqrt(9 + 4)`
= `sqrt 13` ...(iii)
PS = `sqrt((-2 - 2)^2 + (-5 - 1)^2)`
= `sqrt((-4)^2 + (-6)^2)`
= `sqrt(16 + 36)`
= `sqrt52`
= `sqrt(2 xx 2 xx 13)`
= 2`sqrt 13` ...(iv)
In ▢PQRS,
PQ = RS ...[From (i) and (iii)]
QR = PS ...[From (ii) and (iv)]
∴ ▢PQRS is a parallelogram ...(A quadrilateral is a parallelogram if its opposite sides are equal)
By distance formula,
PR = `sqrt((-5 - 2)^2 + (-3 - 1)^2)`
= `sqrt((-7)^2 + (-4)^2)`
= `sqrt(49 +16)`
= `sqrt 65` ...(v)
QS = `sqrt([-2 - (-1)]^2 + (-5 - 3)^2)`
= `sqrt((-7)^2 + (-4)^2)`
= `sqrt(1 + 64)`
= `sqrt 65` ...(vi)
In parallelogram PQRS,
PQ = QS ...[From (v) and (vi)]
∴ ▢PQRS is a rectangle ...(A parallelogram is a rectangle, if its diagonals are equal.)
P(2, 1), Q(–1, 3), R(–5, –3) and S(–2, –5) are the vertices of a rectangle.
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संबंधित प्रश्न
If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b). Prove that bx = ay.
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
Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear.
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)
Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3)
Find the distance of the following points from the origin:
B(–5, 5)
Find the distance between the following pair of points.
L(5, –8), M(–7, –3)
Find the distance of the following point from the origin :
(5 , 12)
Prove that the points (0,3) , (4,3) and `(2, 3+2sqrt 3)` are the vertices of an equilateral triangle.
ABC is an equilateral triangle . If the coordinates of A and B are (1 , 1) and (- 1 , -1) , find the coordinates of C.
Find the co-ordinates of points on the x-axis which are at a distance of 17 units from the point (11, -8).
Find the coordinates of the points on the y-axis, which are at a distance of 10 units from the point (-8, 4).
A point A is at a distance of `sqrt(10)` unit from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.
Given A = (3, 1) and B = (0, y - 1). Find y if AB = 5.
Find distance of point A(6, 8) from origin.
Show that the points (2, 0), (–2, 0) and (0, 2) are vertices of a triangle. State the type of triangle with reason.
The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the figure 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:
What are the coordinates of the position of a player Q such that his distance from K is twice his distance from E and K, Q and E are collinear?
