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the coordinates of the points A and B are (-4,5) and (-5,-4) respectively.A' is reflection image of A with respect to y-axis.B is rotated anticlockwise about the origin O through 90     to B'

Guest Jan 15, 2015

#2
+20025
+5

the coordinates of the points A and B are (-4,5) and (-5,-4) respectively.A' is reflection image of A with respect to y-axis.B is rotated anticlockwise about the origin O through 90     to B'

$$\small{\text{ The vector dot product \vec{A'B} *\vec{AB'}=0, if \vec{A'B} perpendicular \vec{AB'} . }}\\ \small{\text{ We calculate: \vec{A'B} = \vec{A'}-\vec{B}=\left(\begin{array}{c}4\\5\end{array}\right)-\left(\begin{array}{c}-5\\-4\end{array}\right)=\left(\begin{array}{c}4-(-5)\\5-(-4)\end{array}\right)}=\left(\begin{array}{c}9\\9\end{array}\right) }}\\ \small{\text{ and calculate: \vec{AB'} = \vec{A}-\vec{B'}=\left(\begin{array}{c}-4\\5\end{array}\right)-\left(\begin{array}{c}4\\-5\end{array}\right)=\left(\begin{array}{c}-4-4)\\5-(-5)\end{array}\right)}= \left(\begin{array}{c}-8\\10\end{array}\right)  }}\\ \small{\text{ \vec{A'B} *\vec{AB'} =\left(\begin{array}{c}9\\9\end{array}\right) *\left(\begin{array}{c}-8\\10\end{array}\right) =9*(-8)+9*10=-72+90=18  }}\\ \small{\text{ 18 \ne 0  so  \vec{A'B}  not perpendicular to  \vec{AB'} }}$$

heureka  Jan 15, 2015
#1
+17745
+5

Since A = (-4,5) and A' is the reflection image wrt y-axis, A' = (4, 5).

Since B = (-5,-4) and B' is the 90° anticlockwise rotation, B' = (4,-5)

To determine whether or not A'B is perpendicular to AB', find the slopes of A'B and AB'. If they are negative reciprocals, then the lines will be perpendicular.

Is this enough help?

geno3141  Jan 15, 2015
#2
+20025
+5

the coordinates of the points A and B are (-4,5) and (-5,-4) respectively.A' is reflection image of A with respect to y-axis.B is rotated anticlockwise about the origin O through 90     to B'

$$\small{\text{ The vector dot product \vec{A'B} *\vec{AB'}=0, if \vec{A'B} perpendicular \vec{AB'} . }}\\ \small{\text{ We calculate: \vec{A'B} = \vec{A'}-\vec{B}=\left(\begin{array}{c}4\\5\end{array}\right)-\left(\begin{array}{c}-5\\-4\end{array}\right)=\left(\begin{array}{c}4-(-5)\\5-(-4)\end{array}\right)}=\left(\begin{array}{c}9\\9\end{array}\right) }}\\ \small{\text{ and calculate: \vec{AB'} = \vec{A}-\vec{B'}=\left(\begin{array}{c}-4\\5\end{array}\right)-\left(\begin{array}{c}4\\-5\end{array}\right)=\left(\begin{array}{c}-4-4)\\5-(-5)\end{array}\right)}= \left(\begin{array}{c}-8\\10\end{array}\right)  }}\\ \small{\text{ \vec{A'B} *\vec{AB'} =\left(\begin{array}{c}9\\9\end{array}\right) *\left(\begin{array}{c}-8\\10\end{array}\right) =9*(-8)+9*10=-72+90=18  }}\\ \small{\text{ 18 \ne 0  so  \vec{A'B}  not perpendicular to  \vec{AB'} }}$$