Question

If A (3, 0), B (4, 5), C (-1, 4) and D (-2, -1) be four points in a plane, show that ABCD is

a rhombus but not a square​

Answer 1

Answer:

find the length nof each sides if sides are equal then it may be a square or rhombus

to confirm that also find diagonals

if diagonals are not equal then it must be a square

Answer 2

Answer:

Step-by-step explanation:

Let write the equation of line going through AB

Slope $m_{AB}$ =  $\frac{5-0}{4-3}$ = 5

y - 5 = 5(x - 4) ==> y = 5x - 15

Let write the equation of line going through CD

Slope $m_{CD}$ =  $\frac{-1-4}{-2-(-1)}$ = 5  ==> AB ║ CD

y - 4 = 5(x - (- 1)) ==> y = 5x + 9

Let write the equation of line going through AD

Slope $m_{AD}$ = $\frac{-1-0}{-2-3}$ = $\frac{1}{5}$

y - 0 =  $\frac{1}{5}$ (x - 3) ==> y =  $\frac{1}{5}$ x - $\frac{3}{5}$

Let write the equation of line going through BC

Slope $m_{BC}$ = $\frac{4-5}{-1-4}$ = $\frac{1}{5}$  ==> AD ║ BC

y - 4 = $\frac{1}{5}$ (x - (- 1))  ==> y = $\frac{1}{5}$ x + $\frac{21}{5}$

Lines are parallel, but not perpendicular. Perpendicular lines have slopes that are the opposite of the reciprocal of each other.

So far we we prove that ABCD a parallelogram.

AB = √[(5 - 0)² + (4 - 3)²] = √26

BC = √[(5 - 4)² + (4 - (- 1))²] = √26

AB = BC

Thus, we can conclude that ABCD is a rhombus.