Kyle has drawn triangle $ABC$ on a grid. Holly has started to draw an identical triangle $DEF$. We need to find the coordinates of point $F$. The coordinates of the vertices of triangle $ABC$ are $A(3, 5)$, $B(7, 6)$, and $C(4, 8)$. The coordinates of $D$ and $E$ are $D(2, 1)$ and $E(7, 1)$. Since triangle $DEF$ is identical to triangle $ABC$, we can determine the position of $F$ relative to $D$ and $E$ based on the position of $A$ relative to $B$ and $C$.

GeometryCoordinate GeometryVectorsTransformationsTriangles
2025/4/4

1. Problem Description

Kyle has drawn triangle ABCABC on a grid. Holly has started to draw an identical triangle DEFDEF. We need to find the coordinates of point FF. The coordinates of the vertices of triangle ABCABC are A(3,5)A(3, 5), B(7,6)B(7, 6), and C(4,8)C(4, 8). The coordinates of DD and EE are D(2,1)D(2, 1) and E(7,1)E(7, 1). Since triangle DEFDEF is identical to triangle ABCABC, we can determine the position of FF relative to DD and EE based on the position of AA relative to BB and CC.

2. Solution Steps

First, find the vectors BA\vec{BA} and BC\vec{BC}.
BA=AB=(3,5)(7,6)=(37,56)=(4,1)\vec{BA} = A - B = (3, 5) - (7, 6) = (3-7, 5-6) = (-4, -1).
BE=ED=(7,1)(2,1)=(72,11)=(5,0)\vec{BE} = E - D = (7, 1) - (2, 1) = (7-2, 1-1) = (5, 0).
Next, determine what transformation maps BB to EE. B(7,6)B(7, 6) and E(7,1)E(7, 1).
The x-coordinate is the same but the y coordinate changes from 66 to 11. It seems the identical triangle DEFDEF must be a horizontal translation and then a shift down. The corresponding points are as follows: AA to DD, BB to EE and CC to FF.
Therefore, vector AB\vec{AB} is equal to vector DE\vec{DE}.
BA=(7,6)(3,5)=(4,1)B - A = (7, 6) - (3, 5) = (4, 1).
ED=(7,1)(2,1)=(5,0)E - D = (7, 1) - (2, 1) = (5, 0).
Since the vector DEDE should match vector ABAB it seems.
The difference in the vector DE\vec{DE} and AB\vec{AB} indicates a slight stretch and rotation. Since they are supposed to be "identical" this would mean that point AA would be the new point DD point BB would be point EE and point CC would be point FF.
So AC=CA=(4,8)(3,5)=(1,3)\vec{AC} = C - A = (4, 8) - (3, 5) = (1, 3).
We need to add this vector to DD.
F=D+AC=(2,1)+(1,3)=(3,4)F = D + \vec{AC} = (2, 1) + (1, 3) = (3, 4).

3. Final Answer

(3, 4)

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