What is the best solution for the coordinate geometry questions , find the best and simple way to solve the coordinate geometry questions . The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

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

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Let ABCD is a square, where A(-1,2) and B(3,2). And Point O is the point of intersection of AC and BD

To Find: Coordinate of points B and D.

Step 1: Find distance between A and C and coordinates of point O.We know that, diagonals of a square are equal and bisect each other.

AC = √[(3 + 1)

^{2 }+ (2 – 2)^{2}] = 4Coordinates of O can be calculated as follows:

x = (3 – 1)/2 = 1 and y = (2 + 2)/2 = 2

So, O(1,2)

Step 2: Find the side of the square using Pythagoras theoremLet a be the side of square and AC = 4

From right triangle, ACD,

a = 2√2

Hence, each side of square = 2√2

Step 3: Find coordinates of point DEquate length measure of AD and CD

Say, if coordinate of D are (x

_{1}, y_{1})AD = √[(x

_{1}+ 1)^{2 }+ (y_{1}– 2)^{2}]Squaring both sides,

AD

^{2}= (x_{1}+ 1)^{2 }+ (y_{1}– 2)^{2}Similarly, CD

^{2}= (x_{1}– 3)^{2 }+ (y_{1}– 2)^{2}Since all sides of a square are equal, which means AD = CD

(x

_{1}+ 1)^{2 }+ (y_{1}– 2)^{2}= (x_{1}– 3)^{2 }+ (y_{1}– 2)^{2}x

_{1}^{2}+ 1 + 2x_{1}= x_{1}^{2}+ 9 – 6x_{1}8x

_{1}= 8x

_{1}= 1Value of y

_{1}can be calculated as follows by using the value of x.From step 2: each side of square = 2√2

CD

^{2}= (x_{1}– 3)^{2 }+ (y_{1}– 2)^{2}8 = (1 – 3)

^{2 }+ (y_{1}– 2)^{2}8 = 4 + (y

_{1}– 2)^{2}y

_{1}– 2 = 2y

_{1}= 4Hence, D = (1, 4)

Step 4: Find coordinates of point BFrom line segment, BOD

Coordinates of B can be calculated using coordinates of O; as follows:

Earlier, we had calculated O = (1, 2)

Say B = (x

_{2}, y_{2})For BD;

1 = (x

_{2}+ 1)/2x

_{2 }= 1And 2 = (y

_{2}+ 4)/2=> y

_{2}= 0Therefore, the coordinates of required points are B = (1,0) and D = (1,4)