Question

show that the points A(2,1) , B(5,2), C (6,4) and D(3,3) are the vertices of parallelogram. show that ABCD is not a rectangle​

Answer 1

Answer:

Parallelogram ABCD is not a rectangle.

First let us give names to the points:

A(2,1)

B(5,2)

C(6,4)

D(3,3)

These are the vertices of the Parallelogram.

Solving using distance formula,

Distance Formula:

Now solving all by using this Formula:

AC^{2}

=> (6-2)^{2} + (4-1)^{2}

=> (4)^{2} +(3)^{2}  

=> 16+9 = 25

BC^{2} = (6-5)^{2}  + (4-2)^{2}

=> (1)^{2} +(2)^{2}

=> 1+4 = 5

AB^{2} = (5-2)^{2} +(2-1)^{2}

=> (3)^{2} +(1)^{2}

=> 9+1 = 10

DC^{2}

=> (6-3)^{2} +(4-3)^{2}

=> (3)^{2} +(1)^{2}  

=> 9+1=10

AD^{2}

=> (3-2)^{2} +(3-1)^{2}

=> (1)^{2} +(2)^{2}

=> 1+4= 5

Since BC = AD and DC = AB,

ABCD is a parallelogram.

AB^{2}   + BC^{2}  

= 10+5 = 15

AB^{2}  + BC^{2}  ≠ AC^{2}

∴△ABC is not right angled

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