Question

1. draw a line representing the

ise\ and a line representing the
un\ of the line. state the slope of the line in simplest form.

Explanation:

Step1: Identify two points on the line

Looking at the graph, we can see that the line passes through the points \((-10, 0)\) and \((-1, 6)\) (or we can use the two marked points: let's take \((-1, 6)\) and \((4, 9)\) as another pair). Let's use \((-1, 6)\) and \((4, 9)\).

Step2: Calculate the rise

Rise is the change in \(y\)-coordinates, so \(y_2 - y_1 = 9 - 6 = 3\).

Step3: Calculate the run

Run is the change in \(x\)-coordinates, so \(x_2 - x_1 = 4 - (-1) = 5\) (Wait, no, if we take \((-10, 0)\) and \((-1, 6)\): rise is \(6 - 0 = 6\), run is \(-1 - (-10) = 9\)? Wait, maybe better to use the two points where the right triangle is drawn. The lower point is at \(y = 6\), \(x=-1\) and the upper point is at \(y = 9\), \(x = 4\)? Wait, no, looking at the graph, the horizontal line (run) is from \(x=-1\) to \(x = 4\)? Wait, no, the grid: let's check the \(y\)-values. The line starts at \((-10, 0)\), goes through \((-1, 6)\) (since at \(x=-1\), \(y=6\)) and then at \(x = 4\), \(y=9\)? Wait, the vertical line (rise) is from \(y=6\) to \(y=9\), so rise is \(9 - 6 = 3\). The horizontal line (run) is from \(x=-1\) to \(x = 4\), so run is \(4 - (-1) = 5\)? Wait, no, maybe the two points are \((-1, 6)\) and \((4, 9)\). So rise is \(9 - 6 = 3\), run is \(4 - (-1) = 5\)? Wait, but let's check the slope formula: slope \(m=\frac{\text{rise}}{\text{run}}=\frac{y_2 - y_1}{x_2 - x_1}\).

Wait, another way: the line passes through \((-10, 0)\) and \((-1, 6)\). So \(y_2 - y_1 = 6 - 0 = 6\), \(x_2 - x_1 = -1 - (-10) = 9\). Then slope is \(\frac{6}{9}=\frac{2}{3}\)? Wait, no, that can't be. Wait, maybe I made a mistake. Let's look at the \(y\)-intercept: the line crosses the \(y\)-axis at \(y = 7\)? Wait, the graph shows at \(x=0\), \(y=7\)? Wait, maybe the two points are \((-10, 0)\) and \((0, 7)\)? Let's check: from \(x=-10\) (y=0) to \(x=0\) (y=7). Then rise is \(7 - 0 = 7\), run is \(0 - (-10) = 10\)? No, that doesn't match. Wait, the right triangle drawn: the horizontal line (run) is from \(x=-1\) (y=6) to \(x = 4\) (y=6), so run is \(4 - (-1) = 5\). The vertical line (rise) is from \(y=6\) to \(y=9\), so rise is \(9 - 6 = 3\). So slope is \(\frac{3}{5}\)? Wait, no, that doesn't seem right. Wait, let's check the two points: when \(x=-10\), \(y=0\); when \(x=-1\), \(y=6\). So the change in \(y\) is \(6 - 0 = 6\), change in \(x\) is \(-1 - (-10) = 9\). So slope is \(\frac{6}{9}=\frac{2}{3}\). Wait, but the right triangle: maybe the points are \((-1, 6)\) and \((4, 9)\). Then \(y\) changes by \(3\), \(x\) changes by \(5\), so slope \(\frac{3}{5}\)? No, that's conflicting. Wait, let's count the grid squares. From \((-10, 0)\) to \((-1, 6)\): moving 9 units right (from \(x=-10\) to \(x=-1\)) and 6 units up (from \(y=0\) to \(y=6\)). So rise is 6, run is 9. Simplify \(\frac{6}{9}=\frac{2}{3}\). Wait, but the right triangle in the graph: the vertical line is from \(y=6\) to \(y=9\) (3 units up) and horizontal line from \(x=-1\) to \(x=4\) (5 units right). But that would be a different segment. Wait, maybe the two points are \((-1, 6)\) and \((4, 9)\). Then slope is \(\frac{9 - 6}{4 - (-1)}=\frac{3}{5}\). But that contradicts the other calculation. Wait, maybe I misread the graph. Let's look again: the line starts at \((-10, 0)\), goes through \((-1, 6)\) (since at \(x=-1\), \(y=6\)) and then at \(x=4\), \(y=9\) (since from \(x=-1\) to \(x=4\) is 5 units right, and \(y\) goes from 6 to 9, 3 units up). But also, from \(x=-10\) to \(x=-1\) is 9 units right, \(y\) from 0 to 6, 6 units up. So both segments have the same slope? Wait, \(\frac{6}{9}=\frac{2}{3}\) and \(\frac{3}{5}\) are different. That means I made a mistake. Wait, no, the line's slope should be consistent. Let's check the \(y\)-intercept. At \(x=0\), what's \(y\)? The line passes through \((-10, 0)\) and \((-1, 6)\). The slope is \(\frac{6 - 0}{-1 - (-10)}=\frac{6}{9}=\frac{2}{3}\). So the equation is \(y - 0=\frac{2}{3}(x + 10)\), so \(y=\frac{2}{3}x+\frac{20}{3}\approx6.666\). But at \(x=0\), \(y=\frac{20}{3}\approx6.666\), but the graph shows at \(x=0\), \(y=7\)? Maybe the points are \((-1, 6)\) and \((4, 9)\): slope \(\frac{3}{5}=0.6\), and \(\frac{2}{3}\approx0.666\). Hmm, maybe the right triangle is drawn from \((-1, 6)\) to \((4, 9)\), so rise is 3, run is 5, so slope is \(\frac{3}{5}\)? Wait, no, let's check the grid. Each square is 1 unit. So from \(x=-1\) (column -1) to \(x=4\) (column 4) is 5 units (4 - (-1) = 5). From \(y=6\) (row 6) to \(y=9\) (row 9) is 3 units (9 - 6 = 3). So rise is 3, run is 5, so slope is \(\frac{3}{5}\). Wait, but then from \((-10, 0)\) to \((-1, 6)\): \(x\) changes by 9, \(y\) changes by 6, \(\frac{6}{9}=\frac{2}{3}\). There's a contradiction. Wait, maybe the line is not passing through \((-10, 0)\) correctly. Wait, the leftmost point is at \(x=-10\), \(y=0\) (since it's on the \(x\)-axis). Then at \(x=-1\), \(y=6\) (since it's on the horizontal line at \(y=6\)). Then the slope is \(\frac{6 - 0}{-1 - (-10)}=\frac{6}{9}=\frac{2}{3}\). So maybe the right triangle drawn is incorrect, or I misread the points. Wait, the vertical line in the graph is from \(y=6\) to \(y=9\), so that's a rise of 3, and the horizontal line is from \(x=-1\) to \(x=4\), run of 5. But then the slope would be \(\frac{3}{5}\), but that doesn't match the other segment. Wait, maybe the two points are \((-1, 6)\) and \((4, 9)\), so slope is \(\frac{3}{5}\). Alternatively, maybe the line passes through \((-10, 0)\) and \((4, 9)\): rise is \(9 - 0 = 9\), run is \(4 - (-10) = 14\)? No, that's not. Wait, let's use the slope formula correctly. The slope of a line is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take two clear points:

Point 1: \((-10, 0)\) (left end, on \(x\)-axis)

Point 2: \((-1, 6)\) (where the horizontal line starts)

So \(y_2 - y_1 = 6 - 0 = 6\)

\(x_2 - x_1 = -1 - (-10) = 9\)

Thus, slope \(m=\frac{6}{9}=\frac{2}{3}\). Wait, but the vertical line in the graph is from \(y=6\) to \(y=9\) (rise 3) and horizontal from \(x=-1\) to \(x=4\) (run 5). But \(\frac{3}{5}\) is not equal to \(\frac{2}{3}\). That means I made a mistake in identifying the points. Wait, maybe the line passes through \((-1, 6)\) and \((4, 9)\), so slope is \(\frac{3}{5}\), and also through \((-10, 0)\) and \((-1, 6)\) with slope \(\frac{6}{9}=\frac{2}{3}\). That can't be, so I must have misread the points. Wait, let's check the \(y\)-intercept. If the line passes through \((-1, 6)\) and has slope \(\frac{2}{3}\), then the equation is \(y - 6=\frac{2}{3}(x + 1)\), so \(y=\frac{2}{3}x+\frac{2}{3}+6=\frac{2}{3}x+\frac{20}{3}\approx6.666\). But at \(x=0\), \(y=\frac{20}{3}\approx6.666\), but the graph shows a point at \(y=7\) when \(x=0\)? Maybe the graph is a bit off, or I misread the coordinates. Alternatively, maybe the two points are \((-1, 6)\) and \((4, 9)\), so rise is 3, run is 5, slope is \(\frac{3}{5}\). Wait, but let's count the grid squares again. From \(x=-1\) to \(x=4\) is 5 units (4 - (-1) = 5). From \(y=6\) to \(y=9\) is 3 units (9 - 6 = 3). So slope is \(\frac{3}{5}\). But then from \((-10, 0)\) to \((-1, 6)\): \(x\) changes by 9, \(y\) changes by 6, \(\frac{6}{9}=\frac{2}{3}\). There's a discrepancy. Wait, maybe the line is not straight? No, it's a straight line. So perhaps the correct points are \((-1, 6)\) and \((4, 9)\), so slope is \(\frac{3}{5}\). Wait, no, let's check with \(x=0\). If slope is \(\frac{3}{5}\), then from \((-1, 6)\), moving 1 unit right (run=1), rise= \(\frac{3}{5}\), so \(y=6+\frac{3}{5}=6.6\), which is close to 7? Maybe the graph is approximate. Alternatively, maybe the rise is 2 and run is 3? Wait, no. Let's look at the initial point: at \(x=-10\), \(y=0\); at \(x=-7\), \(y=2\); \(x=-4\), \(y=4\); \(x=-1\), \(y=6\); \(x=2\), \(y=8\); \(x=5\), \(y=10\)? Wait, no, the graph shows at \(x=4\), \(y=9\). Wait, maybe the correct slope is \(\frac{2}{3}\). Let's recalculate:

From \((-10, 0)\) to \((-1, 6)\):

\(y_2 - y_1 = 6 - 0 = 6\)

\(x_2 - x_1 = -1 - (-10) = 9\)

\(m = \frac{6}{9} = \frac{2}{3}\)

Yes, that's simplified. So rise is 6, run is 9, slope is \(\frac{2}{3}\). Wait, but the right triangle in the graph: maybe the vertical line is from \(y=6\) to \(y=8\) (rise 2) and horizontal from \(x=-1\) to \(x=2\) (run 3), so slope \(\frac{2}{3}\). Ah, that makes sense! So I misread the upper point. The upper point is at \(y=8\), not 9. So from \(y=6\) to \(y=8\) is rise 2, from \(x=-1\) to \(x=2\) is run 3. So slope is \(\frac{2}{3}\). That matches the first calculation. So the rise is 2, run is 3, slope is \(\frac{2}{3}\).

Wait, let's check: from \((-1, 6)\) to \((2, 8)\): \(y\) changes by 2, \(x\) changes by 3, slope \(\frac{2}{3}\). From \((-10, 0)\) to \((-1, 6)\): \(x\) changes by 9, \(y\) changes by 6, \(\frac{6}{9}=\frac{2}{3}\). Yes, that's consistent. So the rise is 2, run is 3, slope is \(\frac{2}{3}\).

Step4: Correct the rise and run

So the rise (change in \(y\)) is \(8 - 6 = 2\) (if the upper point is at \(y=8\), \(x=2\)) and run (change in \(x\)) is \(2 - (-1) = 3\). Thus, slope \(m=\frac{\text{rise}}{\text{run}}=\frac{2}{3}\).

Answer:

The slope of the line is \(\boxed{\dfrac{2}{3}}\)