Question

the table shows the coordinates of the vertices of \\( \triangle abc \\) and its image.
\\( \

$$\begin{array} { | c | c | } { \\hline \\triangle a b c } & { \\triangle a ^ { \\prime } b ^ { \\prime } c ^ { \\prime } } \\\\ \\hline a ( 0,0 ) & { a ^ { \\prime } ( 0,10 ) } \\\\ \\hline b ( 10, - 20 ) & { b ^ { \\prime } ( 10, - 10 ) } \\\\ \\hline c ( - 10, - 20 ) & { c ^ { \\prime } ( - 10, - 10 ) } \\\\ \\hline \\end{array}$$

\\)
which translation maps \\( \triangle a b c \\) to its image?
10 units down
10 units up
10 units to the right
10 units to the left

Explanation:

Step1: Analyze the change in y - coordinates

For point \(A(0,0)\) and \(A'(0,10)\), the y - coordinate changes from \(y = 0\) to \(y=10\). For point \(B(10,-20)\) and \(B'(10,-10)\), the y - coordinate changes from \(y=-20\) to \(y = - 10\). For point \(C(-10,-20)\) and \(C'(-10,-10)\), the y - coordinate changes from \(y=-20\) to \(y=-10\).
The formula for a vertical translation is \((x,y)\to(x,y + k)\). Here, if \((x,y)\) is the original point and \((x,y')\) is the translated point, then \(y'=y + k\).
Substituting values:
For \(A\): \(10=0 + k\), so \(k = 10\).
For \(B\): \(-10=-20 + k\), solving for \(k\) gives \(k=-10+20 = 10\).
For \(C\): \(-10=-20 + k\), solving for \(k\) gives \(k=-10 + 20=10\).

Step2: Determine the direction of translation

Since \(k = 10>0\), and in the coordinate system, when \(k>0\) in the translation \((x,y)\to(x,y + k)\), the figure is translated \(k\) units up.

Answer:

10 units up