Question

(−4,−5) (−5,−4)
(−3,−3) (−3,−3)
(−2,−1) (−1,−2)
(−1,1) (1,−1)
(0,3) (3,0)
(1,5) (5,1)
(2,7) (7,2)
(3,9) (9,3)

Explanation:

Step1: Analyze the table

The table has pairs of points like \((-4, -5)\) and \((-5, -4)\), \((-3, -3)\) and \((-3, -3)\), \((-2, -1)\) and \((-1, -2)\), etc. Notice that for most pairs, if we have a point \((x, y)\) in the first column, the second column has \((y, x)\). This suggests a reflection over the line \(y = x\).

Step2: Analyze the graph

The graph has two lines with yellow points and some red points. Let's check the points from the table. For example, \((0, 3)\) and \((3, 0)\) should be reflections over \(y = x\). The line \(y = x\) has a slope of 1 and passes through the origin. Looking at the yellow points, one line seems to have points like \((0, 3)\), \((1, 5)\), \((2, 7)\), \((3, 9)\) which has a slope of \(\frac{5 - 3}{1 - 0} = 2\) (wait, actually \(\frac{y - 3}{x - 0}\), for \((1,5)\): \(\frac{5 - 3}{1 - 0} = 2\), so equation \(y = 2x + 3\)? Wait no, maybe I made a mistake. Wait the other line has points like \((3, 0)\), \((5, 1)\), \((7, 2)\), \((9, 3)\). Let's check the slope: \(\frac{1 - 0}{5 - 3} = \frac{1}{2}\), so equation \(y=\frac{1}{2}x - \frac{3}{2}\)? Wait no, maybe the key is the reflection over \(y = x\). The line \(y = x\) is the line where \(x = y\). The points \((x, y)\) and \((y, x)\) are symmetric over \(y = x\). So the two lines in the graph should be symmetric with respect to \(y = x\). Let's check the yellow points: one line has \((0, 3)\), \((1, 5)\), \((2, 7)\), \((3, 9)\) (let's call this Line 1) and the other has \((3, 0)\), \((5, 1)\), \((7, 2)\), \((9, 3)\) (Line 2). Let's see if Line 2 is the reflection of Line 1 over \(y = x\). For a point \((x, y)\) on Line 1, \((y, x)\) should be on Line 2. For \((0, 3)\), \((3, 0)\) is on Line 2. For \((1, 5)\), \((5, 1)\) is on Line 2. For \((2, 7)\), \((7, 2)\) is on Line 2. For \((3, 9)\), \((9, 3)\) is on Line 2. Perfect! So Line 1 has points \((x, 2x + 3)\) (wait \((0,3)\): \(y = 2(0)+3 = 3\), \((1,5)\): \(2(1)+3 = 5\), yes! So Line 1: \(y = 2x + 3\). Line 2: let's take \((3,0)\): if we solve \(x = 2y + 3\) (since it's the reflection over \(y = x\), we swap \(x\) and \(y\) in Line 1's equation: \(x = 2y + 3\), then solve for \(y\): \(2y = x - 3\), \(y=\frac{1}{2}x - \frac{3}{2}\). Let's check \((3,0)\): \(y=\frac{1}{2}(3)-\frac{3}{2}=0\), correct. \((5,1)\): \(\frac{1}{2}(5)-\frac{3}{2}=\frac{5 - 3}{2}=1\), correct. So Line 2 is the reflection of Line 1 over \(y = x\).

Now, the red points: let's see if they are on the reflection. For example, the red point near \((0,3)\): maybe \((0,3)\) is yellow, and the red point is maybe a mistake or a test point. Wait the question is probably to identify the relationship, but since the user didn't specify the question, but from the table and graph, it's about coordinate geometry, specifically reflection over the line \(y = x\).

But since the user didn't ask a specific question, maybe the task is to identify the transformation or the equation of the lines. Wait maybe the question is to find the equation of the lines or identify the symmetry. Let's assume the question is to find the equation of the two lines.

For Line 1 (with points \((0, 3)\), \((1, 5)\), \((2, 7)\), \((3, 9)\)):

Step1: Find slope

Slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{5 - 3}{1 - 0}=2\)

Step2: Find y-intercept

Using point-slope form \(y - y_1 = m(x - x_1)\) with \((0,3)\):

\(y - 3 = 2(x - 0)\)

\(y = 2x + 3\)

For Line 2 (with points \((3, 0)\), \((5, 1)\), \((7, 2)\), \((9, 3)\)):

Step1: Find slope

Slope \(m=\frac{1 - 0}{5 - 3}=\frac{1}{2}\)

Step2: Find y-intercept

Using point-slope form with \((3,0)\):

\(y - 0=\frac{1}{2}(x - 3)\)

\(y=\frac{1}{2}x - \frac{3}{2}\)

Alternatively, since Line 2 is the reflection of Line 1 over \(y = x\), we can swap \(x\) and \(y\) in Line 1's equation:

\(x = 2y + 3\)

Solve for \(y\):

\(2y = x - 3\)

\(y=\frac{1}{2}x - \frac{3}{2}\), which matches.

So the two lines are \(y = 2x + 3\) and \(y=\frac{1}{2}x - \frac{3}{2}\), and they are symmetric with respect to the line \(y = x\) (since swapping \(x\) and \(y\) in one equation gives the other).

Answer:

The two lines are \(y = 2x + 3\) and \(y=\frac{1}{2}x - \frac{3}{2}\), and they are symmetric with respect to the line \(y = x\) (reflection over \(y = x\)).