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प्रश्न
Show that points P(2, –2), Q(7, 3), R(11, –1) and S (6, –6) are vertices of a parallelogram.
उत्तर
The given points are P(2, –2), Q(7, 3), R(11, –1) and S(6, –6).
`"Distance between" = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`
By distance formula,
PQ = `sqrt((7 - 2)^2 + [3 - ( - 2)]^2)`
∴ PQ = `sqrt((7 - 2)^2 + (3 + 2)^2)`
∴ PQ = `sqrt((5)^2 + (5)^2)`
∴ PQ = `sqrt(25 + 25)`
∴ PQ = `sqrt(50)`
∴ PQ = `sqrt(25 × 2)`
∴ PQ = `5sqrt(2)` ...(1)
QR = `sqrt((11 - 7)^2 + (-1 - 3)^2)`
∴ QR = `sqrt((4)^2 + (-4)^2)`
∴ QR = `sqrt(16 + 16)`
∴ QR = `sqrt(32)`
∴ QR = `sqrt(16 × 2)`
∴ QR = `4sqrt(2)` ...(2)
RS = `sqrt((6 - 11)^2 + [-6 - (- 1)]^2)`
∴ RS = `sqrt((- 5)^2 + (-6 + 1)^2)`
∴ RS = `sqrt((- 5)^2 + (-5)^2)`
∴ RS = `sqrt(25 + 25)`
∴ RS = `sqrt(50)`
∴ RS = `sqrt(25 × 2)`
∴ RS = `5sqrt(2)` ...(3)
PS = `sqrt((6 - 2)^2 + [-6 - (- 2)]^2)`
∴ PS = `sqrt((6 - 2)^2 + [-6 + 2]^2)`
∴ PS = `sqrt((4)^2 + (-4)^2)`
∴ PS = `sqrt(16 + 16)`
∴ PS = `sqrt(32)`
∴ PS = `sqrt(16 × 2)`
∴ PS = `4sqrt(2)` ...(4)
In □ PQRS,
PQ = RS ...[From (1) and (3)]
QR = PS ...[From (2) and (4)]
A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.
Checking for slopes,
Slope of a line between two points (x1, y1) and (x2, y2) is
m = `(y_2 - y_1)/(x_2 - x_1)`
Slope PQ = `(7 - 2)/[3 - (- 2)] = 1`
Slope QR = `(11 - 7)/[- 1 - 3] = - 1`
Slope RS = `(6 - 11)/[- 6 - (- 1)] = 1`
Slope SP = `(6 - 2)/[- 6 - (- 2)] = - 1`
As PQ = RS and their slope = 1 And QR = SP and their slope = -1.
∴ □ PQRS is parallelogram.
∴ P, Q, R, and S are vertices of a parallelogram.
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