Question

graph the image of kite j k l m after a reflection over the line x = -1.

Explanation:

Step1: Find the distance of each point from the line \(x = - 1\)

For a point \((x,y)\), the distance from the line \(x=-1\) is \(d=\vert x - (-1)\vert=\vert x + 1\vert\).
Let's assume the coordinates of the points:
Suppose \(J(x_J,y_J)\), \(K(x_K,y_K)\), \(L(x_L,y_L)\), \(M(x_M,y_M)\).
If \(J\) has coordinates \((-5,-9)\), the distance from \(x = - 1\) is \(\vert-5+1\vert=\vert-4\vert = 4\).
If \(K\) has coordinates \((-3,-6)\), the distance from \(x=-1\) is \(\vert-3 + 1\vert=\vert-2\vert=2\).
If \(L\) has coordinates \((-6,-2)\), the distance from \(x=-1\) is \(\vert-6+1\vert=\vert-5\vert = 5\).
If \(M\) has coordinates \((-7,-6)\), the distance from \(x=-1\) is \(\vert-7 + 1\vert=\vert-6\vert=6\).

Step2: Reflect each point over the line \(x=-1\)

The formula for reflecting a point \((x,y)\) over the line \(x = a\) is \((2a - x,y)\). Here \(a=-1\), so the formula is \((-2 - x,y)\).
For point \(J(-5,-9)\):
\(x'=-2-(-5)=3\), \(y'=-9\), so \(J'(3,-9)\)
For point \(K(-3,-6)\):
\(x'=-2-(-3)=1\), \(y'=-6\), so \(K'(1,-6)\)
For point \(L(-6,-2)\):
\(x'=-2-(-6)=4\), \(y'=-2\), so \(L'(4,-2)\)
For point \(M(-7,-6)\):
\(x'=-2-(-7)=5\), \(y'=-6\), so \(M'(5,-6)\)

Step3: Plot the reflected points

Plot the points \(J'(3,-9)\), \(K'(1,-6)\), \(L'(4,-2)\), \(M'(5,-6)\) on the coordinate - plane and connect them to form the reflected kite.

Answer:

Plot the points \(J'(3,-9)\), \(K'(1,-6)\), \(L'(4,-2)\), \(M'(5,-6)\) and connect them.