Question

reflect the figure over the line $y = -3$. plot all of the points of the reflected figure. you may click a plotted point to delete it. answer attempt 1 out of 2

Explanation:

Step1: Find the distance between each point and the line \(y = - 3\)

Let's assume the original points of the triangle are \(A(-5,-7)\), \(B(-2,-5)\), \(C(1,-7)\).
For a point \((x,y)\), the distance \(d\) from the point \((x,y)\) to the line \(y=-3\) is \(d=\vert y - (-3)\vert=\vert y + 3\vert\).
For point \(A(-5,-7)\): \(d_A=\vert-7+3\vert = 4\)
For point \(B(-2,-5)\): \(d_B=\vert-5 + 3\vert=2\)
For point \(C(1,-7)\): \(d_C=\vert-7+3\vert = 4\)

Step2: Calculate the \(y\) - coordinate of the reflected points

The formula for reflecting a point \((x,y)\) over the line \(y = k\) is \((x,2k - y)\). Here \(k=-3\), so the formula is \((x,-6 - y)\)
For point \(A(-5,-7)\): \(y\) - coordinate of reflected point \(A'\) is \(-6-(-7)=1\), so \(A'(-5,1)\)
For point \(B(-2,-5)\): \(y\) - coordinate of reflected point \(B'\) is \(-6-(-5)=-1\), so \(B'(-2,-1)\)
For point \(C(1,-7)\): \(y\) - coordinate of reflected point \(C'\) is \(-6-(-7)=1\), so \(C'(1,1)\)

Answer:

Plot the points \((-5,1)\), \((-2,-1)\) and \((1,1)\)