Question

(−2, 4), (4, 8), (10, 7) and (11, –5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.

Answer 1

Let the given points be A(−2, 4), B(4, 8), C(10, 7) and D(11, −5).

Let P, Q, R and S be the mid-points of AB, BC, CD and DA respectively.

Co-ordinates of P are `((-2 + 4)/2, (4 + 8)/2) = (1, 6)`

Co-ordinates of Q are `((4 + 10)/2, (8 + 7)/2) = (7, 15/2)`

Co-ordinates of R are `((10 + 11)/2, (7 - 5)/2) = (21/2, 1)`

Co-ordinates of S are `((- 2 + 11)/2, (4 - 5)/2) = (9/2, (-1)/2)`

Slope of PQ = `(15/2 - 6)/(7 - 1) = ((15 - 12)/2)/6 = 3/12 = 1/4`

Slope of RS = `((-1)/2 - 1)/(9/2 - 21/2) = ((- 1 - 2)/2)/((9 - 21)/2) = (- 3)/(- 12) = 1/4`

Since, slope of PQ = Slope of RS, PQ || RS.

Slope of QR = `(1 - 15/2)/(21/2 - 7) = ((2 - 15)/2)/((21 - 14)/2) = -13/7`

Slope of SP = `(6 + 1/2)/(1 - 9/2) = ((12 + 1)/2)/((2 - 9)/2) = -13/7`

Since, slope of QR = Slope of SP, QR || SP.

Hence, PQRS is a parallelogram.