Question

r(-2, -5), δxyz: x(-11, 4), y(0, -6), z(-1, 3)
xy = \sqrt{(0 + 11)^2 + (-6 - 4)^2}

Explanation:

Step1: Identify the distance formula

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For points \(X(-11, 4)\) and \(Y(0, -6)\), \(x_1=-11,y_1 = 4,x_2 = 0,y_2=-6\).

Step2: Substitute values into the formula

Substitute into the distance formula: \(XY=\sqrt{(0 - (-11))^2+(-6 - 4)^2}=\sqrt{(0 + 11)^2+(-10)^2}=\sqrt{11^2+(- 10)^2}\)

Step3: Calculate the squares

Calculate \(11^2 = 121\) and \((-10)^2=100\). Then \(XY=\sqrt{121 + 100}\)

Step4: Add the numbers inside the square root

\(121+100 = 221\), so \(XY=\sqrt{221}\approx14.87\) (if a decimal approximation is needed) or leave it as \(\sqrt{221}\)

Answer:

If we simplify the expression \(\sqrt{(0 + 11)^2+(-6 - 4)^2}\), we get \(\sqrt{121 + 100}=\sqrt{221}\approx14.87\) (or \(\sqrt{221}\) exactly). So the length of \(XY\) is \(\sqrt{221}\) (or approximately \(14.87\)).