Question

13.) shantell wrote a reflection over the line ( y = x - 10 ) will carry the rectangle onto itself. however maria wrote a reflection over ( y = - x + 4 ) will carry the rectangle onto itself. are both students correct? graph both lines and explain why.

Explanation:

Step1: Recall the property of reflection

A reflection over a line maps a figure onto itself if the line is a line of symmetry of the figure.

Step2: Check for line of symmetry

For a rectangle, the lines \(y = x - 10\) and \(y=-x + 4\) need to be checked if they are lines of symmetry.
Let's assume the rectangle has vertices \(A\), \(K\), \(T\), \(E\).
For a line \(y = mx + c\) to be a line of symmetry of a figure, for every point \((x,y)\) on one side of the line, there must be a corresponding point \((x',y')\) on the other side such that the line is the perpendicular bisector of the segment joining \((x,y)\) and \((x',y')\).
Let's take two opposite vertices (say \(A\) and \(T\)) and check their mid - point.
The mid - point of two points \((x_1,y_1)\) and \((x_2,y_2)\) is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\).
If we assume \(A=(3,5)\) and \(T=( - 3,11)\) (by observing the graph), the mid - point of \(A\) and \(T\) is \((\frac{3+( - 3)}{2},\frac{5 + 11}{2})=(0,8)\)
For the line \(y=x - 10\), when \(x = 0\), \(y=-10
eq8\). So \(y=x - 10\) is not a line of symmetry.
For the line \(y=-x + 4\), when \(x = 0\), \(y = 4
eq8\). So \(y=-x + 4\) is not a line of symmetry.

Answer:

No, neither Shantell nor Maria is correct. Because neither \(y=x - 10\) nor \(y=-x + 4\) is a line of symmetry of the given rectangle.