Question

15. Show that (-3, 2), (-5, -5), (2, -3) and
(4, 4) are the vertices of a rhombus.​

Answer 1

Answer:

Step 1 - Find the distance of the points P(-3, 2), Q(-5, -5), R(2, -3) and  S(4, 4).

Using d=√((x₂-x₁)²+(y₂-y₁)²), We find:

• PQ=√53 units.

$d = \sqrt {(-5 - (-3))^2 + (-5 - 2)^2}$

$d = \sqrt {(-2)^2 + (-7)^2}$

$d = \sqrt {4 + 49}$

$d=\sqrt{54}$

• QR=√53 units.

$d = \sqrt {(2 - (-5))^2 + (-3 - (-5))^2}$

$d = \sqrt {(7)^2 + (2)^2}$

$d=\sqrt{49+4}$

$d=\sqrt{54}$

• RS=√53 units.

$d = \sqrt {(4 - 2)^2 + (4 - (-3))^2}$

$d = \sqrt {(2)^2 + (7)^2}$

$d = \sqrt {4 + 49}$

$d=\sqrt{54}$

• SP=√53 units.

$d = \sqrt {(-3 - 4)^2 + (2 - 4)^2}$

$d = \sqrt {(-7)^2 + (-2)^2}$

$d = \sqrt {49 + 4}$

$d=\sqrt{54$

THUS, all sides are EQUAL. We know that if a quadrilateral has equal sides, it is a rhombus. You might wonder that square also has equal sides. But SQUARE IS ALSO A RHOMBUS.

$\therefore \framebox{PQRS is a rhombus}$

$\framebox[1.1\width]{\huge HOPE THIS HELPS :D}$