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Question 4 The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices. Solution

O is the point of intersection of AC and BD. Coordinates of O can be calculated as follows: x=3−12=1 y=2+22=2 Now, AC can be calculated as follows: Diagonals of a square are equal and bisect each other. ⇒AC=√(3+1)2+(2−2)2 =√16=4 Hence, sides of square = 2√2 (Using Pythagoras theorem) Let coordinates of point D be (x1,y1) AD=2√2=√(x1+1)2+(y1−2)2 ⇒8=(x1+1)2+(y1−2)2 CD=2√2=√(x1−3)2+(y1−2)2 ⇒8=(x1−3)2+(y1−2)2 From these equations, it is clear that; (x1+1)2+(y1−2)2 =(x1−3)2+(y1−2)2 ⇒(x1+1)2=(x1−3)2 ⇒x12+2x+1=x12−6x+9 ⇒2x+1=−6x+9 ⇒8x=8 ⇒x=1 Value of y1 can be calculated as follows by using the value of x. CD=2√2=√(x1−3)2+(y1−2)2 ⇒8=(1−3)2+(y1−2)2 ⇒4+(y1−2)2=8 ⇒y1−2=2 ⇒y1=4 Hence, D = (1, 4) Coordinates of B can be calculated using coordinates of O; as follows: Earlier, we had calculated O = (1, 2) For BD, using section formula, 1=x+12 ⇒x+1=2 ⇒x=1 2=y+42 ⇒y+4=4 ⇒y=0 Hence, B = (1, 0) and D = (1, 4)Mathematics

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