Question

If A(5,y), B(1,5), C(2,1) and D(6,2) are the vertices of a square, then the value of y is

Answer 1

Answer:

The value of y is 6.

Step-by-step explanation:

Since if A(5,y), B(1,5), C(2,1) and D(6,2) are the vertices of a square, then the value of y is possible to compute graphically.

I suggest you to draw it in a paper, and you will get a three vertices.

The fact that we are leading with a square is very important because it gives us the information that its vertices have a relation of proportionality between each pair of vertices, so if you draw the vertices you will repair that from point C to point D the vertice moves 1 unit up and 4 to the right. So the same will happen between point A and B.

Hence, the value of y is 6.

Answer 2

To find the value of y,if A(5,y), B(1,5), C(2,1) and D(6,2) are the vertices of a square,use the property of square,i. e.

Each side of square are equal.

Apply distance formula AB =DC

$AB = \sqrt{ {(5 - 1)}^{2} + {(y - 5)}^{2} } \\ \\ AB = \sqrt{16 + {y}^{2} + 25 - 10y } \\ \\ AB = \sqrt{ {y}^{2} - 10y + 41 }$
$CD = \sqrt{ {(2 - 6)}^{2} + {(1 - 2)}^{2} } \\ \\ CD = \sqrt{16 +1 } \\ \\ CD = \sqrt{17 }$
Since both the distance are equal ,equate AB=DC

${y}^{2} - 10y + 41 = 17 \\ \\ {y}^{2} - 10y + 24 = 0 \\ \\ {y}^{2} - 6y - 4y + 24 = 0 \\ \\ y(y - 6) - 4(y - 6) = 0 \\ \\ (y - 6)(y -4 ) = 0 \\ \\ y = 6 \\ \\ y = 4$
So, value of y can be 4 and 6.