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 ) \\)

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 \(d=x - (-1)=x + 1\).
For \(J(2,-1)\), \(d_J=2+1 = 3\).
For \(K(4,-5)\), \(d_K=4 + 1=5\).
For \(L(3,1)\), \(d_L=3+1 = 4\).

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 formula becomes \((-2 - x,y)\).
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 \(L'(-5,1)\).

Answer:

The coordinates of \(\triangle JKL\) are \(J(2,-1)\), \(K(4,-5)\), \(L(3,1)\) and the coordinates of its image after reflection over the line \(x = - 1\) are \(J'(-4,-1)\), \(K'(-6,-5)\), \(L'(-5,1)\). Plot these points on the coordinate - plane to graph \(\triangle JKL\) and its image.