Question

The two opposite vertices of a square are(-1,2)qnd(3,-2) find the coordinate of other two vertices

Answer 1

Step-by-step explanation:

Given The two opposite vertices of a square are(-1,2) and (3,2) find the coordinate of other two vertices

• Given two opposite vertices of a square are (-1, 2) and (3,  2)
• In a square all sides are equal.
• Let (x,y) be the vertex.
• Therefore PQ = RS
• Now we have the distance formula
• PQ = √(x + 1)^2 + (y – 2)^2
• RS = √(x – 3)^2 + (y - 2)^2
• Now PQ = RS
• So  √(x + 1)^2 + (y – 2)^2 = √(x – 3)^2 + (y - 2)^2
• Taking out the square root we get
• (x + 1)^2 + (y – 2)^2 = (x – 3)^2 + (y - 2)^2
• (x + 1)^2 = (x – 3)^2
• Or x^2 + 2x + 1 = x^2 – 6x + 9  
• Or x^2 + 2x + 1 – x^2 + 6x – 9 = 0
• 8x – 8 = 0
• Or x = 1
• Now by joining the diagonal PR and using pythagoras theorem we get
• PR^2 = PQ^2 + PR^2
• [(√3 + 1)^2 + (2 – 2)]^2)^2 = [(√(x + 1)^2 + (y – 2)^2]^2 + [√(x – 3)^2 + (y – 2)^2]^2
• 4^2 + 0 = (x + 1)^2 + (y – 2)^2 + (x – 3)^2 + (y – 2)^2
• 16 = x^2 + 2x + 1 + y^2 – 4y + 4 + x^2 – 6x + 9 + y^2 - 4y + 4
• 16 = 2x^2 + 2y^2 – 4x – 8y + 18
• 2x^2 + 2y^2 – 4x – 8y + 2 = 0
• Putting x= 1 we get
• 2(1)^2 + 2y^2 – 4(1) – 8y + 2 = 0
• 2 + 2y^2 – 4 – 8y + 2 = 0
• 2y^2 – 8y = 0
• 2y(y – 4) = 0
• Or y = 0 , y – 4 = 0
• So y = 0 and y = 4

Therefore the other two vertices are (1,0) and (1,4)

Reference link will be

https://brainly.in/question/8081429