Show that points P(2, -2), Q(7, 3), R(11, -1) and S (6, -6) are vertices of a parallelogram.
Answer:-
[Distance between” = sqrt{((x_2 – x_1)^2 + (y_2 – y_1)^2)}]
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{t(50)} ]
[∴ RS = sqrt{t(25 × 2)} ]
[∴ RS = 5 sqrt{(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 = 4 sqrt{(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.
Chapter 5. Co-ordinate Geometry – Practice Set 5.1 (Page 107)