Question

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. (EXPLAIN WORK)

Answer 1

A(2,2)

Explanation:

Let the vertex A has coordinates
`(x_A,y_A)`

Vectors AB and AB' are perpendicular, then

`\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`

Vectors AC and AC' are perpendicular, then

`\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`

Now, solve the system of two equations:

`\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.`

Subtract these two equations:

`5x_A-y_A-8=0\Rightarrow y_A=5x_A-8`

Substitute it into the first equation:

`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)}=(60\pm8)/(2\cdot 13)=(34)/(13),2`

Then

`y_{A_(1,2)}=5\cdot (34)/(13)-8 \text{ or } 5\cdot 2-8\\ \\=(66)/(13)\text{ or } 2`

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)