Question

problem #4 - unit 1 (medium)
find the coordinates of the vertices of each figure after a reflection across the y-axis
k(1, -1), n(4, 0), q(4, -4) k( , ), n( , ), q( , )
problem #5 - unit 1 (medium)
reflect the following image over the
x-axis and label appropriately.

Explanation:

Problem #4 Solution:

Step1: Recall reflection over y - axis rule

The rule for reflecting a point \((x,y)\) across the \(y\) - axis is \((x,y)\to(-x,y)\).

Step2: Find \(K'\)

For point \(K(1, - 1)\), applying the rule: \(x = 1\), so \(-x=-1\), \(y=-1\). So \(K'(-1,-1)\).

Step3: Find \(N'\)

For point \(N(4,0)\), applying the rule: \(x = 4\), so \(-x=-4\), \(y = 0\). So \(N'(-4,0)\).

Step4: Find \(Q'\)

For point \(Q(4,-4)\), applying the rule: \(x = 4\), so \(-x=-4\), \(y=-4\). So \(Q'(-4,-4)\).

To reflect a point \((x,y)\) across the \(x\) - axis, we use the rule \((x,y)\to(x,-y)\).

1. First, identify the coordinates of the vertices of the original triangle (let's say the vertices are \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(C(x_3,y_3)\) from the graph).
2. Then, for each vertex, apply the reflection rule: \(A'(x_1,-y_1)\), \(B'(x_2,-y_2)\), \(C'(x_3,-y_3)\).
3. Plot these new points \(A'\), \(B'\), \(C'\) and connect them to form the reflected triangle.

For example, if one of the vertices of the original triangle (from the graph) is \((a,b)\), its reflection over the \(x\) - axis will be \((a, - b)\). After finding the coordinates of all three vertices using this rule, we can label the reflected triangle with these new coordinates.

(Note: Since the exact coordinates of the original vertices are not clearly discernible from the provided image, the above is a general method to solve the reflection over the \(x\) - axis problem. If the coordinates of the original vertices were known, we could substitute them into the rule \((x,y)\to(x,-y)\) to get the reflected vertices.)

Answer:

\(K'(-1,-1)\), \(N'(-4,0)\), \(Q'(-4,-4)\)

Problem #5 Solution (General Explanation as Coordinates of Original Points are not Clearly Visible from the Image):