Question

1. (02.03 mc)

what coordinate for f would make triangle abc and triangle def congruent? (1 point)

(-2,4)
(-2,3)
(-3,3)
(-3,4)

Explanation:

Step1: Determine the coordinates of points in \(\triangle ABC\)

• \(A(1,0)\), \(B(-2,2)\), \(C(2,3)\)

Step2: Analyze the transformation for congruence

• For \(\triangle ABC\) and \(\triangle DEF\) to be congruent, assume a translation.
• \(D(-1,0)\) (from the graph), \(A(1,0)\) to \(D(-1,0)\) is a translation of \(x - 2\).
• \(B(-2,2)\) after \(x - 2\) is \(x=-2-2=-4\) (incorrect approach, better use side - length and shape matching)
• Calculate the length of \(AC\): \(\sqrt{(2 - 1)^{2}+(3 - 0)^{2}}=\sqrt{1 + 9}=\sqrt{10}\)
• Calculate the length of \(AB\): \(\sqrt{(-2 - 1)^{2}+(2 - 0)^{2}}=\sqrt{9 + 4}=\sqrt{13}\)
• Calculate the length of \(BC\): \(\sqrt{(2+2)^{2}+(3 - 2)^{2}}=\sqrt{16 + 1}=\sqrt{17}\)
• Now, check the options. If we consider the vertical and horizontal distances.
• The \(y\) - coordinate of \(C\) is \(3\), for congruence, the \(y\) - coordinate of \(F\) should be \(3\) (eliminates \((-2,4)\) and \((-3,4)\))
• The horizontal distance from \(A\) to \(B\) is \(|-2 - 1| = 3\), from \(D(-1,0)\) if we assume a similar horizontal spread.
• If we consider the position relative to \(D\). The \(x\) - coordinate: \(D(-1,0)\), for a congruent triangle, if we assume a reflection or translation.
• Another way: Count the units. From \(A\) to \(B\) is \(3\) units left and \(2\) units up. From \(D(-1,0)\), if we go \(3\) units left (since \(A(1,0)\) to \(D(-1,0)\) is \(2\) units left, but using the shape).
• By visual inspection (since \(C\) is at \((2,3)\), for congruence, moving the triangle so that \(D\) is a vertex. The \(x\) - coordinate of \(F\) should be \(-3\) (because of the left - right spread) and \(y\) - coordinate \(3\)

Answer:

\((-3,3)\)