Question

lesson 5 | session 2 develop performing sequences of transformations involving dilations read and try to solve the problem below. noe learns to program graphics on a computer. he draws △abc. first he reflects it across the x - axis to form △abc. then he dilates △abc using a scale factor of 1/2 with the center of dilation at the origin. the final image is △abc. what are the coordinates of the vertices of △abc? try it math toolkit graph paper, tracing paper, transparency sheets discuss it ask: why did you choose that strategy to find the coordinates of △abc? share: i knew... so i...

Explanation:

Step1: Recall reflection rule

When reflecting a point $(x,y)$ across the $x -$axis, the transformation is $(x,y)\to(x, - y)$. Let the coordinates of vertices of $\triangle ABC$ be $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$. After reflection across the $x -$axis, the vertices of $\triangle A'B'C'$ are $A'(x_1,-y_1)$, $B'(x_2,-y_2)$, $C'(x_3,-y_3)$.

Step2: Recall dilation rule

The rule for dilation with a scale factor $k=\frac{1}{2}$ and center of dilation at the origin $(0,0)$ is $(x,y)\to(kx,ky)$. For the vertices of $\triangle A'B'C'$ with coordinates $A'(x_1,-y_1)$, $B'(x_2,-y_2)$, $C'(x_3,-y_3)$, after dilation, the vertices of $\triangle A''B''C''$ are $A''(\frac{1}{2}x_1,-\frac{1}{2}y_1)$, $B''(\frac{1}{2}x_2,-\frac{1}{2}y_2)$, $C''(\frac{1}{2}x_3,-\frac{1}{2}y_3)$. You need to substitute the actual coordinates of $A$, $B$, $C$ from the graph into these formulas to get the exact coordinates of $\triangle A''B''C''$.

Answer:

The coordinates of $\triangle A''B''C''$ are obtained by first reflecting the coordinates of $\triangle ABC$ across the $x -$axis and then dilating the resulting coordinates by a scale factor of $\frac{1}{2}$ with the center of dilation at the origin. If the coordinates of $A=(x_A,y_A)$, $B=(x_B,y_B)$, $C=(x_C,y_C)$, then $A''=(\frac{1}{2}x_A,-\frac{1}{2}y_A)$, $B''=(\frac{1}{2}x_B,-\frac{1}{2}y_B)$, $C''=(\frac{1}{2}x_C,-\frac{1}{2}y_C)$.