Question

graph △jkl and its image after a reflection in the line y = 1.
j(2, 4), k(-4, -2), l(-1, 0)
polygon undo redo × reset

Explanation:

Step1: Find the distance from each point to the line \(y = 1\)

For a point \((x,y)\), the distance to the line \(y = 1\) is \(d=\vert y - 1\vert\).

• For \(J(2,4)\): \(d_J=\vert4 - 1\vert=3\)
• For \(K(-4,-2)\): \(d_K=\vert-2 - 1\vert = 3\)
• For \(L(-1,0)\): \(d_L=\vert0 - 1\vert=1\)

Step2: Reflect the points across the line \(y = 1\)

When reflecting a point \((x,y)\) over the line \(y = 1\), the \(x\)-coordinate remains the same, and the new \(y\)-coordinate is \(y'=1-(y - 1)\)

• For \(J(2,4)\): \(y'_J=1-(4 - 1)=-2\), so \(J'(2,-2)\)
• For \(K(-4,-2)\): \(y'_K=1-(-2 - 1)=4\), so \(K'(-4,4)\)
• For \(L(-1,0)\): \(y'_L=1-(0 - 1)=2\), so \(L'(-1,2)\)

Step3: Graph \(\triangle JKL\) and \(\triangle J'K'L'\)

• Plot \(J(2,4)\), \(K(-4,-2)\), \(L(-1,0)\) and connect them to form \(\triangle JKL\)
• Plot \(J'(2,-2)\), \(K'(-4,4)\), \(L'(-1,2)\) and connect them to form \(\triangle J'K'L'\)

Answer:

The coordinates of the reflected triangle are \(J'(2,-2)\), \(K'(-4,4)\), \(L'(-1,2)\)