Question

triangle def is congruent to right triangle abc, shown below. if point e has the coordinates (1,5), what could be the coordinates of point f? (-2,1) (-3,2) (1,0) (1,2)

Explanation:

Step1: Find the length of \(BC\)

Coordinates of \(B(-3,0)\) and \(C(0,-3)\). Using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), for \(B(-3,0)\) and \(C(0,-3)\), we have \(d_{BC}=\sqrt{(0 + 3)^2+(-3-0)^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}\). Also, since \(\triangle ABC\) is a right - triangle, \(AB=\sqrt{(0 + 3)^2+(4 - 0)^2}=\sqrt{9+16}=5\) and \(AC=\sqrt{(0 - 0)^2+(-3 - 4)^2}=7\). But since \(\triangle DEF\cong\triangle ABC\) and \(E(1,5)\), if we consider the vertical or horizontal distance (because of congruence and right - triangle property).
If we assume \(EF\) is vertical (similar to \(AC\) or \(AB\) in orientation). The length of \(AB = 5\). If \(E(1,5)\) and we consider a vertical segment (because in right - triangle \(ABC\), \(AB\) is a non - horizontal, non - vertical side, but if we assume a translation or transformation). Wait, another approach: count the units. In \(\triangle ABC\), from \(A(0,4)\) to \(B(-3,0)\) is \(5\) units (using the distance formula as above). From \(E(1,5)\), if we move down \(5\) units (since the length of the corresponding side in \(\triangle ABC\) is \(5\)).
The \(y\) - coordinate of \(E\) is \(5\). If we move down \(5\) units (\(y=5-5 = 0\)) and keep \(x\) - coordinate same (if the side is vertical). But let's check the options.
If we consider the fact that in congruent triangles, corresponding sides are equal. In right - triangle \(ABC\), \(AB = 5\). If \(E(1,5)\) and \(F\) is such that the distance between \(E\) and \(F\) is \(5\).
Using the distance formula for each option:

• For \((-2,1)\): \(d=\sqrt{(1 + 2)^2+(5 - 1)^2}=\sqrt{9 + 16}=5\)
• For \((-3,2)\): \(d=\sqrt{(1+3)^2+(5 - 2)^2}=\sqrt{16 + 9}=5\) (but we need to check the right - triangle property. If \(\triangle DEF\) is right - triangle. If \(E(1,5)\) and \(D\) is a point (not given exactly, but from the figure \(D\) is at \((4,2)\) approximately. Wait, another way: in \(\triangle ABC\), \(AB\) is a side. If \(\triangle DEF\cong\triangle ABC\), and assume \(DE\) corresponds to \(AB\). But since \(E(1,5)\), if we consider the vertical and horizontal differences.

If we assume \(EF\) is vertical. The length of \(AC\) is \(7\) (from \(A(0,4)\) to \(C(0,-3)\)). But no. Wait, count the grid units: \(AB\) is \(5\) units (from \((0,4)\) to \((-3,0)\) is \(5\) units as \(3\) units in \(x\) and \(4\) units in \(y\), \(3 - 4-5\) triangle). For \(E(1,5)\) to \((-2,1)\): \(x\) difference \(1+2 = 3\), \(y\) difference \(5 - 1=4\), so by Pythagorean theorem \(d=\sqrt{3^2+4^2}=5\)

Answer:

\((-2,1)\)