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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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