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5 from unit 1, lesson 9 points a, b, c, and d are vertices of a square.…

Question

5 from unit 1, lesson 9 points a, b, c, and d are vertices of a square. point e is inside the square. explain how to tell whether point e is closer to a, b, c, or d.

Explanation:

Step1: Use the distance formula

If we assume the coordinates of \(A=(x_1,y_1)\), \(B=(x_2,y_2)\), \(C=(x_3,y_3)\), \(D=(x_4,y_4)\) and \(E=(x_E,y_E)\). The distance formula is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Calculate \(d_{EA}=\sqrt{(x_E - x_1)^2+(y_E - y_1)^2}\), \(d_{EB}=\sqrt{(x_E - x_2)^2+(y_E - y_2)^2}\), \(d_{EC}=\sqrt{(x_E - x_3)^2+(y_E - y_3)^2}\), \(d_{ED}=\sqrt{(x_E - x_4)^2+(y_E - y_4)^2}\).

Step2: Compare the distances

Compare the values of \(d_{EA}\), \(d_{EB}\), \(d_{EC}\) and \(d_{ED}\). The smallest value among them will tell which vertex ( \(A\), \(B\), \(C\) or \(D\)) point \(E\) is closest to.

Answer:

Calculate the distances from point \(E\) to \(A\), \(B\), \(C\) and \(D\) using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) and then compare these distances. The vertex corresponding to the smallest distance is the one that point \(E\) is closest to.