Show that points P(2, -2), Q(7, 3), R(11, -1) and S (6, -6) are vertices of a parallelogram.
\[Distance between" = \sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2)}\]
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,
\[PQ = \sqrt{((7 - 2)^2 + [3 - ( - 2)]^2)}\]
\[∴ PQ = \sqrt{((7 - 2)^2 + (3 + 2)^2)}\]
\[∴ PQ = \sqrt{((5)^2 + (5)^2)} \]
\[∴ PQ = \sqrt{(25 + 25)} \]
\[∴ PQ = \sqrt{(50)} \]
\[∴ PQ = \sqrt{(25 × 2)} \]
\[∴ PQ = 5\sqrt{(2) ...(1)} \]
\[QR = \sqrt{((11 - 7)^2 + (-1 - 3)^2)} \]
\[∴ QR = \sqrt{((4)^2 + (-4)^2)} \]
\[∴ QR = \sqrt{(16 + 16)} \]
\[∴ QR = \sqrt{(32)} \]
\[∴ QR = \sqrt{(16 × 2)} \]
\[∴ QR = 4\sqrt{(2) ...(2)} \]
\[RS = \sqrt{((6 - 11)^2 + [-6 - (- 1)]^2)} \]
\[∴ RS = \sqrt{((- 5)^2 + (-6 + 1)^2)} \]
\[∴ RS = \sqrt{((- 5)^2 + (-5)^2)} \]
\[∴ RS = \sqrt{(25 + 25)} \]
\[∴ RS = \sqrt{t(50)} \]
\[∴ RS = \sqrt{t(25 × 2)} \]
\[∴ RS = 5 \sqrt{(2) ...(3)} \]
\[PS = \sqrt{((6 - 2)^2 + [-6 - (- 2)]^2)} \]
\[∴ PS = \sqrt{((6 - 2)^2 + [-6 + 2]^2)} \]
\[∴ PS = \sqrt{((4)^2 + (-4)^2)} \]
\[∴ PS = \sqrt{(16 + 16)} \]
\[∴ PS = \sqrt{(32)} \]
\[∴ PS = \sqrt{(16 × 2)} \]
\[∴ PS = 4 \sqrt{(2) ...(4)} \]
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)