Question

Show that points A(-4,-7), B (-1,2), C(8,5)and D(5,-4) are vertices of a rhombus ABCD.​

Answer 1

Answer:

Let the points are A(−4,−7), B(−1,2), C(8,5) and D(5,−4).

AB = √(−1−(−4))² + (2−(−7))²

= √(3² + 9²)

= √90

= 3√10units.

BC= √(8−(−1))² + (5−2)²

= √(9² + 3²)

= √90

= 3√10units.

CD = √(5−8)² + (−4−(5))²

= √(3² +9²)

= √90

= 3√10units.

DA = √(−4−(5))² + (−7−(−4))²

= √(9² +3²)

= √90

= 3√10units.

Now ,

AC = √(8−(−4))² + (5−(−7))²

= √(12² +12²)

= √144

= 12√2units.

BD = √(5−(−1))² + (−4−2)²

= √(6² + 6²)

= √72

= 6√2units.

Since , AB = CD = BC = DA ; AC ≠ BD

so, it is rhombus. proved.

Answer 2

See the picture attached to get your answer..

Hope it would help you out.