Show that the points 1,7 4,2 -1,-1 and -4,4 are vertice of a square

1.	Show that the points 1,7 4,2 -1,-1 and -4,4 are vertice of a square
Answer» ong>Answer: We know that, diagonals of a square bisect each other. So,if mid POINT of DIAGONAL AC is equal to mid point of diagonal BD,then ABCD is a parallelogram. Then if it is proved that the diagonals are equal then ABCD is a rectangle. And then,if it is proved that the adjacent sides are equal,then we are confirmed that the GIVEN points are VERTICES of square. Midpoint of AC=1-1/2,7-1/2=0,3 Midpoint of BD=4-4/2,2+4/2=0,3 Since,the midpoints are equal. Therefore,ABCD is a parallelogram. AC=((-1-1)^2+(-1-7)^2)^1/2=(4+64)^1/2=68^1/2 BD=((-4-4)^2+(4-2)^2)^1/2=(64+4)^1/2=68^1/2 Since,the diagonals are equal,ABCD is a rectangle. AB=((4-1)^2+(2-7)^2)^1/2=(9+25)^1/2=34^1/2. BC=((-1-4)^2+(-1-2)^2)^1/2=(25+9)^1/2=34^1/2. Since,the adjacent sides are equal,we are confirmed that the given points are vertices of square. Hope the answer helps. If it helps,mark the brainliest. Thank you.

Answer»

ong>Answer:

We know that,

diagonals of a square bisect each other.

So,if mid POINT of DIAGONAL AC is equal to mid point of diagonal BD,then ABCD is a parallelogram.

Then if it is proved that the diagonals are equal then ABCD is a rectangle.

And then,if it is proved that the adjacent sides are equal,then we are confirmed that the GIVEN points are VERTICES of square.

Midpoint of AC=1-1/2,7-1/2=0,3

Midpoint of BD=4-4/2,2+4/2=0,3

Since,the midpoints are equal.

Therefore,ABCD is a parallelogram.

AC=((-1-1)^2+(-1-7)^2)^1/2=(4+64)^1/2=68^1/2

BD=((-4-4)^2+(4-2)^2)^1/2=(64+4)^1/2=68^1/2

Since,the diagonals are equal,ABCD is a rectangle.

AB=((4-1)^2+(2-7)^2)^1/2=(9+25)^1/2=34^1/2.

BC=((-1-4)^2+(-1-2)^2)^1/2=(25+9)^1/2=34^1/2.

Since,the adjacent sides are equal,we are confirmed that the given points are vertices of square.

Hope the answer helps.

If it helps,mark the brainliest.

Thank you.

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