Find the other two vertices of a square with one vertex (0, 0) and another vertex (4, 2). can you find another answer?

It is given that two vertices of square are (0,0) and (4,2).

Now the problem is that you haven't given that whether these two vertices are adjacent vertices or opposite vertices of the square.

1. By Supposing that these two are adjacent vertices of Square

The third vertex will be at (-4,2) which lies in third quadrant.

Suppose the coordinate of fourth vertex be (x,y).

Mid point of line joining (4,2) and (-4,2) is{ [4+(-4)]/2,(2+2)/2} is (0,2).

Mid point of line joining (x,y) and (0,0) is (x/2,y/2).

Since diagonals of square bisect each other,

∵ x/2=0

⇒x=0

and

y/2=2

⇒y=4

So, The Coordinate of  fourth vertex is (0,4).

Now coming back to second condition if these are two opposite vertex of Square.

Let the third coordinate be (a,b).

Length of diagonal=

Now,let side of Square be A.

Then length of Diagonal of square =√2 A

⇒√2 A=2√5

⇒A =√10

As third vertex is (a,b).

Using distance formula

a² + b²=10  -------------(1)

(a-4)²+ (b-2)²=10  --------------(2)

Solving expression (1) and (2), we get

⇒a²+ b²=(a-4)² +(b-2)²

⇒2a + b =5

⇒b=5-2a

Putting the value of b in (1),we get

⇒a² +(5-2a)²=10

⇒a²+25+4a²-20a =10

⇒5a²-20a+15=0

⇒a² - 4a + 3=0

Splitting the middle term,we get

⇒(a-3)(a-1)=0

⇒a=3  ∧  a=1

we get b=5-2×1=3 and b=5-2×3=5-6=-1

So,the other vertex are (1,3) and(3,-1).

the 3rd option is right

step-by-step explanation:

answer:

a=9

step-by-step explanation: lmk if you need a explanation