Question

if a(2 -1) b(3 4) c(-2 3) d(-3 -2) be four points in a plane show that ABCD is a rhombus

Answer 1

Answer:

ABCD is a rhombus

Step-by-step explanation:

A(2 ,-1), B(3, 4) C(-2,3) D(-3,-2)

$AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$AB=\sqrt{(2-3)^2+(-1-4)^2}$

$AB=\sqrt{1+25}$

$\bf{AB=\sqrt{26}}$

$BC=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

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

$BC=\sqrt{25+1}$

$\bf{BC=\sqrt{26}}$

$CD=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

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

$CD=\sqrt{1+25}$

$\bf{CD=\sqrt{26}}$

$AD=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$AD=\sqrt{(2+3)^2+(-1+2)^2}$

$AD=\sqrt{25+1}$

$\bf{AD=\sqrt{26}}$

$\implies\:AB=BC=CD=AD$

All sides are equal

But,

$AB^2+BC^2\neq\:AC^2$

$\angle{B}\neq90$

$\implies\:\text{ABCD is not a square}$

Hence, ABCD is a rhombus

Answer 2

ABCD is a rhombus

A(2 ,-1), B(3, 4) C(-2,3) D(-3,-2)

All sides are equal

But,

Hence, ABCD is a rhombus

Step-by-step explanation: