Question

graph \\( \triangle j k l \\) and its image after a reflection in the line \\( x = - 1 \\).
\\( j ( 2, - 1 ), k ( 4, - 5 ), l ( 3, 1 ) \\)
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9

Explanation:

Step1: Find the distance between each point and the line \(x = - 1\)

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

• For \(J(2,-1)\): \(d_J=\vert2+1\vert = 3\)
• For \(K(4,-5)\): \(d_K=\vert4 + 1\vert=5\)
• For \(L(3,1)\): \(d_L=\vert3+1\vert = 4\)

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

The formula for reflecting a point \((x,y)\) over the line \(x = a\) is \((2a - x,y)\). Here \(a=-1\), so the reflected point \((x',y')=(2\times(-1)-x,y)=(-2 - x,y)\)

• For \(J(2,-1)\): \(x'=-2-2=-4\), \(y'=-1\), so \(J'(-4,-1)\)
• For \(K(4,-5)\): \(x'=-2 - 4=-6\), \(y'=-5\), so \(K'(-6,-5)\)
• For \(L(3,1)\): \(x'=-2-3=-5\), \(y' = 1\), so \(L'(-5,1)\)

Step3: Graph the original and reflected triangles

Plot the points \(J(2,-1)\), \(K(4,-5)\), \(L(3,1)\) for \(\triangle JKL\) and \(J'(-4,-1)\), \(K'(-6,-5)\), \(L'(-5,1)\) for the reflected \(\triangle J'K'L'\) on the coordinate - plane.

Answer:

The coordinates of the reflected triangle \(\triangle J'K'L'\) are \(J'(-4,-1)\), \(K'(-6,-5)\), \(L'(-5,1)\)