Hushar Mulga
@Rohit
Spread the love

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,

\[PQ = \sqrt{((7 - 2)^2 + [3 - ( - 2)]^2)}\]

\[∴ PQ = \sqrt{((7 - 2)^2 + (3 + 2)^2)}\]

\[∴ PQ = \sqrt{((5)^2 + (5)^2)} \]

\[∴ PQ = \sqrt{(25 + 25)} \]

\[∴ PQ = \sqrt{(50)} \]

\[∴ PQ = \sqrt{(25 × 2)} \]

\[∴ PQ = 5\sqrt{(2)        ...(1)} \]

\[QR = \sqrt{((11 - 7)^2 + (-1 - 3)^2)} \]

\[∴ QR = \sqrt{((4)^2 + (-4)^2)} \]

\[∴ QR = \sqrt{(16 + 16)} \]

\[∴ QR = \sqrt{(32)} \]

\[∴ QR = \sqrt{(16 × 2)} \]

\[∴ QR = 4\sqrt{(2)      ...(2)} \]

\[RS = \sqrt{((6 - 11)^2 + [-6 - (- 1)]^2)} \]

\[∴ RS = \sqrt{((- 5)^2 + (-6 + 1)^2)} \]

\[∴ RS = \sqrt{((- 5)^2 + (-5)^2)} \]

\[∴ RS = \sqrt{(25 + 25)} \]

\[∴ RS = \sqrt{t(50)} \]

\[∴ RS = \sqrt{t(25 × 2)} \]

\[∴ RS = 5 \sqrt{(2)        ...(3)} \]

\[PS = \sqrt{((6 - 2)^2 + [-6 - (- 2)]^2)} \]

\[∴ PS = \sqrt{((6 - 2)^2 + [-6 + 2]^2)} \]

\[∴ PS = \sqrt{((4)^2 + (-4)^2)} \]

\[∴ PS = \sqrt{(16 + 16)} \]

\[∴ PS = \sqrt{(32)} \]

\[∴ PS = \sqrt{(16 × 2)} \]

\[∴ PS = 4 \sqrt{(2)      ...(4)} \]

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)