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Question
Show that points P(1, –2), Q(5, 2), R(3, –1), S(–1, –5) are the vertices of a parallelogram.
Solution
Let, P(1, −2) = (x1, y1); Q(5, 2) = (x2, y2); R(3, −1) = (x3, y3) and S(−1, −5) = (x4, y4).
By distance formula,
PQ = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`
= `sqrt((5 - 1)^2 + [2 - (-2)]^2)`
= `sqrt((4)^2 + (4)^2)`
= `sqrt(16 + 16)`
= `sqrt32`
= `sqrt(2 xx 2 xx 2 xx 2 xx 2)`
= `4sqrt2` ...(1)
QR = `sqrt((x_3 - x_2)^2 + (y_3 - y_2)^2)`
= `sqrt((3 - 5)^2 + (-1 - 2)^2`
= `sqrt((-2)^2 + (-3)^2)`
= `sqrt(4 + 9)`
= `sqrt13` ...(2)
RS = `sqrt((x_4 - x_3)^2 + (y_4 - y_3)^2)`
= `sqrt((-1 - 3)^2 + [-5 - (-1)]^2)`
= `sqrt((-4)^2 + (-4)^2)`
= `sqrt(16 + 16)`
= `sqrt32`
= `sqrt(2 xx 2 xx 2 xx 2 xx 2)`
= `4sqrt2` ...(3)
PS = `sqrt((x_4 - x_1)^2 + (y_4 - y_1)^2)`
= `sqrt((-1 - 1)^2 + [-5 - (-2)]^2)`
= `sqrt((-2)^2 + (-5 + 2)^2)`
= `sqrt((-2)^2 + (-3)^2)`
= `sqrt(4 + 9)`
= `sqrt13` ...(4)
In `square`PQRS,
PQ = RS ...[From (1) and (3)]
QR = PS ...[From (2) and (4)]
`square`PQRS is a parallelogram. ...(A quadrilateral is a parallelogram if both the pairs of its opposite sides are congruent.)
Points P(1, –2), Q(5, 2), R(3, –1), S(–1, –5) are the vertices of a parallelogram.
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