Question

prove that the point (-4,4), (2,-2),(5,1)and (-1,7) taken in order are the vertices of a rectangle ​

Answer 1

Answer:

Hece we know that opp. side of rectangle are equal then AB=CD,BC=AD

Hence it is a rectangle

Answer 2

The point (-4,4), (2,-2), (5,1) and (-1,7) taken in order are the vertices of a rectangle ​

Step-by-step explanation:

Given: A(-4,4), B(2,-2), C(5,1), D (-1,7) are four points taken in order

To prove: ABCD is a rectangle

as we know the distance formula

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

Therefore the distance between A(-4,4), B(2,-2)

$AB=\sqrt{(2-(-4))^2+(-2-4)^2} = \sqrt{36+36} =\sqrt{36(1+1)} =6\sqrt{2}$

The distance between B(2,-2), C(5,1)

$BC=\sqrt{(5-2)^2+(1-(-2))^2} = \sqrt{9+9} =\sqrt{9(1+1)} =3\sqrt{2}$

The distance between C(5,1) and D(-1,7)

$CD=\sqrt{(-1-5)^2+(7-1)^2} = \sqrt{36+36} =\sqrt{36(1+1)} =6\sqrt{2}$

and the distance between A(-4,4) ,D(-1,7)

$AD=\sqrt{(-1-(-4))^2+(7-4)^2} = \sqrt{9+9} =\sqrt{9(1+1)} =3\sqrt{2}$

Now AB=CD and BC=AD

as opposite sides of a quadrilateral are equal then it  is a parallelogram

Now finding the length of diagonal A(-4,4) , C(5,1)

the length of diagonal is given by

$AC=\sqrt{(5-(-4))^2+(1-4)^2} = \sqrt{81+9} =\sqrt{90}$

Now as we can see

AB²+BC²=(6√2)²+(3√2)²= 72+18= 90 = AC²

Therefore by converse of Pythagoras theorem

If he sum of square of the other two sides of a triangle is equal to square of a side then triangle must be right angle triangle.

Therefore ∠B=90°

Hence ABCD is a rectangle

#Learn more

Prove that the points A(2,3) , B(-2,2).C(-1,-2) and D(3,-1) are the vertices of the square ABCD. also find the length of the diagonal.

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