Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and coordinates B’ and C’ of it’s image are (–2, 2) and (1, 4). What is the coordinate of vertex A.
Answers
Step-by-step explanation:
Let the vertex A has coordinates (x_A,y_A)(x
A
,y
A
)
Vectors AB and AB' are perpendicular, then
\begin{gathered}\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\end{gathered}
AB
=(2−x
A
,6−y
A
)
AB
′
=(−2−x
A
,2−y
A
)
AB
⊥
AB
′
⇒
AB
⋅
AB
′
=0⇒(2−x
A
)(−2−x
A
)+(6−y
A
)(2−y
A
)=0
Vectors AC and AC' are perpendicular, then
\begin{gathered}\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{gathered}
AC
=(4−x
A
,3−y
A
)
AC
′
=(1−x
A
,4−y
A
)
AC
⊥
AC
′
⇒
AC
⋅
AC
′
=0⇒(4−x
A
)(1−x
A
)+(3−y
A
)(4−y
A
)=0
Now, solve the system of two equations:
\begin{gathered}\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.\end{gathered}
⎩
⎪
⎨
⎪
⎧
(2−x
A
)(−2−x
A
)+(6−y
A
)(2−y
A
)=0
(4−x
A
)(1−x
A
)+(3−y
A
)(4−y
A
)=0
⎩
⎪
⎨
⎪
⎧
−4−2x
A
+2x
A
+x
A
2
+12−6y
A
−2y
A
+y
A
2
=0
4−4x
A
−x
A
+x
A
2
+12−3y
A
−4y
A
+y
A
2
=0
⎩
⎪
⎨
⎪
⎧
x
A
2
+y
A
2
−8y
A
+8=0
x
A
2
+y
A
2
−5x
A
−7y
A
+16=0
Subtract these two equations:
5x_A-y_A-8=0\Rightarrow y_A=5x_A-85x
A
−y
A
−8=0⇒y
A
=5x
A
−8
Substitute it into the first equation:
\begin{gathered}x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2\end{gathered}
x
A
2
+(5x
A
−8)
2
−8(5x
A
−8)+8=0
x
A
2
+25x
A
2
−80x
A
+64−40x
A
+64+8=0
26x
A
2
−120x
A
+136=0
13x
A
2
−60x
A
+68=0
D=(−60)
2
−4⋅13⋅68=3600−3536=64
x
A
1,2
=
2⋅13
60±8
=
13
34
,2
Then
\begin{gathered}y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2\end{gathered}
y
A
1,2
=5⋅
13
34
−8 or 5⋅2−8
=
13
66
or 2
Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)
