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प्रश्न
Show that the points (0, –1), (8, 3), (6, 7) and (– 2, 3) are vertices of a rectangle.
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उत्तर
Let the points be P(0, –1), Q(8, 3), R(6, 7), S(–2, 3)
Distance between two points= `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
∴ By distance formula,
d(P, Q) = `sqrt((8 - 0)^2 + [3 - (-1)]^2`
= `sqrt((8 - 0)^2 + (3 + 1)^2`
= `sqrt(8^2 + 4^2)`
= `sqrt(64 + 16)`
= `sqrt(80)` ......(i)
d(Q, R) = `sqrt((6 - 8)^2 + (7 - 3)^2`
= `sqrt((-2)^2 + (4)^2`
= `sqrt(4 + 16)`
= `sqrt(20)` ......(ii)
d(R, S) = `sqrt([(-2) - 6]^2 + (3 - 7)^2`
= `sqrt((-8)^2 + (-4)^2`
= `sqrt(64 + 16)`
=`sqrt(80)` ......(iii)
d(P, S) = `sqrt([(-2) - 0]^2 + [3 - (-1)^2]`
= `sqrt((-2)^2+ (3+ 1)^2`
= `sqrt((-2)^2 + 4^2`
= `sqrt(4 + 16)`
= `sqrt(20)` ......(iv)
In ▢PQRS,
∴ side PQ = side RS .......[From (i) and (iii)]
side QR = side PS ......[From (ii) and (iv)]
∴ ▢PQRS is a parallelogram ......[A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent]
d(P, R) = `sqrt((6 - 0)^2 + [7 - (-1)]^2`
= `sqrt((6 - 0)^2 + (7 + 1)^2`
= `sqrt(6^2 + 8^2)`
= `sqrt(36 + 64)`
= `sqrt(100)`
= 10 ......(iv)
d(Q, S) = `sqrt([(-2) - 8]^2 + [3 - 3]^2`
= `sqrt((-10)^2 + (0)^2`
= `sqrt(100 + 0)`
= `sqrt(100)`
= 10 ......(vi)
In parallelogram PQRS,
PR = QS .......[From (v) and (vi)]
∴ ▢PQRS is a rectangle. .......[A parallelogram is a rectangle if its diagonals are equal]
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Tharunya was thrilled to know that the football tournament is fixed with a monthly timeframe from 20th July to 20th August 2023 and for the first time in the FIFA Women’s World Cup’s history, two nations host in 10 venues. Her father felt that the game can be better understood if the position of players is represented as points on a coordinate plane. |
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- Check if the Goal keeper G(–3, 5), Sweeper H(3, 1) and Wing-back K(0, 3) fall on a same straight line.
[or]
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