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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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