Question

1. triangle jkl with vertices j(1, -1), k(2, 3), and l(3, -2) in the line x = 4.

Explanation:

To reflect a point \((x, y)\) over the vertical line \(x = a\), we use the formula for reflection over a vertical line: the new \(x\)-coordinate \(x'\) is given by \(x'=2a - x\), and the \(y\)-coordinate remains the same (\(y' = y\)). Here, the line of reflection is \(x = 4\), so \(a = 4\).

Step 1: Find \(J'\)

For point \(J(1,-1)\), we apply the reflection formula.
The \(x\)-coordinate of \(J'\) is \(2\times4 - 1=8 - 1 = 7\).
The \(y\)-coordinate remains \(-1\) (since reflection over a vertical line doesn't change the \(y\)-coordinate).
So, \(J'(7,-1)\).

Step 2: Find \(K'\)

For point \(K(2,3)\), we apply the reflection formula.
The \(x\)-coordinate of \(K'\) is \(2\times4 - 2 = 8 - 2=6\).
The \(y\)-coordinate remains \(3\).
So, \(K'(6,3)\).

Step 3: Find \(L'\)

For point \(L(3,-2)\), we apply the reflection formula.
The \(x\)-coordinate of \(L'\) is \(2\times4 - 3=8 - 3 = 5\).
The \(y\)-coordinate remains \(-2\).
So, \(L'(5,-2)\).

Answer:

\(J'(7, -1)\)
\(K'(6, 3)\)
\(L'(5, -2)\)