Question

△abc is graphed on the coordinate plane. what is the length of each line segment after a dilation of scale factor 2 centered at the origin? ab = units ac = units enter a number like 2 in the box.

Explanation:

Step1: Recall distance - formula for dilation

When a figure is dilated by a scale factor \(k\) centered at the origin, the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) in the original figure and the corresponding points in the dilated figure is related by the fact that the distance in the dilated figure is \(k\) times the distance in the original figure. First, find the lengths of \(AB\) and \(AC\) in the original triangle using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) or by counting grid - squares for right - angled cases.

Step2: Count grid - squares for original lengths

For \(AC\): In the original triangle, \(A=(0, - 5)\) and \(C=(0,-2)\). The length of \(AC\) is \(|-2-(-5)|=3\) units. For \(AB\): Using the Pythagorean theorem. If we consider the right - triangle formed by the horizontal and vertical displacements from \(A\) to \(B\). The horizontal displacement from \(A=(0, - 5)\) to \(B=(3,-2)\) is \(3 - 0=3\) units and the vertical displacement is \(-2-(-5) = 3\) units. Then \(AB=\sqrt{(3 - 0)^2+(-2 + 5)^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}\) units.

Step3: Apply dilation formula

Since the scale factor \(k = 2\), the length of \(AB\) after dilation is \(AB'=k\times AB=2\times3\sqrt{2}=6\sqrt{2}\) units and the length of \(AC\) after dilation is \(AC'=k\times AC=2\times3 = 6\) units.

Answer:

\(AB = 6\sqrt{2}\)
\(AC = 6\)