Answer:
The answer is ""
Step-by-step explanation:
Distance
midpoint height
let
Gradient
Perpendicular Gradient
midpoints
Similar to perpendicular
Now I'll obtain the perpendicular bisector solution
An equation of the center circle radius will be
.
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).
