Question

find an equation for the line below.

Explanation:

Step1: Identify two points on the line

From the graph, we can see two points: \((-2, 5)\) and \((4, -5)\).

Step2: Calculate the slope \(m\)

The formula for slope is \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
Substituting the points \((x_1, y_1) = (-2, 5)\) and \((x_2, y_2) = (4, -5)\):
\(m = \frac{-5 - 5}{4 - (-2)} = \frac{-10}{6} = -\frac{5}{3}\)

Step3: Use point - slope form to find the equation

The point - slope form is \(y - y_1 = m(x - x_1)\). Let's use the point \((0, 2)\) (we can find the y - intercept by looking at the graph where \(x = 0\), \(y=2\)) or we can use one of the other points. Let's use the slope - intercept form \(y=mx + b\), where \(b\) is the y - intercept. We know that when \(x = 0\), from the graph, the line crosses the y - axis at \(y = 2\), so \(b = 2\).
We already found \(m=-\frac{5}{3}\). So the equation of the line in slope - intercept form \(y=mx + b\) is:
\(y=-\frac{5}{3}x+2\)

We can also verify using the two points. Let's take the point \((-2,5)\):
Left - hand side: \(y = 5\)
Right - hand side: \(-\frac{5}{3}\times(-2)+2=\frac{10}{3}+2=\frac{10 + 6}{3}=\frac{16}{3}\)? Wait, no, maybe we made a mistake in identifying the points. Let's re - identify the points. Looking at the graph again, the two blue points: one is at \((-2,5)\) and the other is at \((4, - 5)\)? Wait, when \(x = 0\), the y - value: let's recalculate the slope with correct points. Wait, maybe the two points are \((-2,5)\) and \((4, - 5)\) is wrong. Let's check the grid again. Each grid square is 1 unit. Let's take the point \((-2,5)\) and \((4, - 5)\): the change in \(y\) is \(-5 - 5=-10\), change in \(x\) is \(4-(-2) = 6\), slope is \(-\frac{10}{6}=-\frac{5}{3}\). But when \(x = 0\), let's plug \(x = 0\) into \(y=-\frac{5}{3}x + b\). Using the point \((-2,5)\): \(5=-\frac{5}{3}\times(-2)+b\), \(5=\frac{10}{3}+b\), \(b = 5-\frac{10}{3}=\frac{15 - 10}{3}=\frac{5}{3}\)? No, that's not correct. Wait, maybe the correct points are \((-2,5)\) and \((4, - 5)\) is incorrect. Let's look at the y - intercept. When \(x = 0\), the line passes through \(y = 2\)? Wait, no, let's count the grid. From \((-2,5)\) to \((0,2)\): the change in \(x\) is \(2\), change in \(y\) is \(-3\), so slope \(m=\frac{-3}{2}\)? Wait, I think I misread the points. Let's re - examine the graph.

Wait, the first blue point: \(x=-2\), \(y = 5\) (since from the origin, moving 2 units left on x - axis and 5 units up on y - axis). The second blue point: \(x = 4\), \(y=-5\) (4 units right on x - axis and 5 units down on y - axis). Wait, the distance between \(x=-2\) and \(x = 4\) is \(6\) units, and between \(y = 5\) and \(y=-5\) is \(10\) units. So slope \(m=\frac{-5 - 5}{4-(-2)}=\frac{-10}{6}=-\frac{5}{3}\). Now, let's use the point \((0,2)\) (when \(x = 0\), \(y = 2\)). Let's check if this point lies on the line with slope \(-\frac{5}{3}\) and passing through \((-2,5)\). The equation using point \((-2,5)\) is \(y - 5=-\frac{5}{3}(x + 2)\).
Expanding: \(y-5=-\frac{5}{3}x-\frac{10}{3}\)
\(y=-\frac{5}{3}x-\frac{10}{3}+5\)
\(y=-\frac{5}{3}x-\frac{10}{3}+\frac{15}{3}\)
\(y=-\frac{5}{3}x+\frac{5}{3}\)? No, that's not matching \(y = 2\) when \(x = 0\). I must have misidentified the y - intercept. Let's look at the graph again. When \(x = 0\), the line crosses the y - axis at \(y = 2\)? Wait, no, let's take the two points correctly. Let's take the point \((-2,5)\) and \((1,0)\)? Wait, maybe the graph has a different scale. Wait, the correct way: let's find two points with integer coordinates.

Looking at the graph, the line passes through \((-2, 5)\) and \((4, - 5)\) is wrong. Let's take \((-2, 5)\) and \((1, 0)\): slope \(m=\frac{0 - 5}{1-(-2)}=\frac{-5}{3}\), same as before. When \(x = 0\), \(y=-\frac{5}{3}(0)+b\), using \((-2,5)\): \(5=-\frac{5}{3}\times(-2)+b\), \(5=\frac{10}{3}+b\), \(b = 5-\frac{10}{3}=\frac{5}{3}\approx1.67\), but the graph seems to cross the y - axis at \(y = 2\). Maybe my initial point identification is wrong. Let's look at the two blue dots: one is at \((-2,5)\) and the other is at \((4, - 5)\). Wait, let's calculate the equation using these two points.

Using the two - point form: \(\frac{y - y_1}{y_2 - y_1}=\frac{x - x_1}{x_2 - x_1}\)
\(\frac{y - 5}{-5 - 5}=\frac{x+2}{4 + 2}\)
\(\frac{y - 5}{-10}=\frac{x + 2}{6}\)
Cross - multiply: \(6(y - 5)=-10(x + 2)\)
\(6y-30=-10x - 20\)
\(6y=-10x + 10\)
\(y=-\frac{10}{6}x+\frac{10}{6}\)
\(y=-\frac{5}{3}x+\frac{5}{3}\)

Wait, but when \(x = 0\), \(y=\frac{5}{3}\approx1.67\), which is close to what we saw. Maybe the graph's y - axis is a bit off. Alternatively, maybe the two points are \((-2,5)\) and \((3, - 5)\)? No, let's check the grid again. Each small square is 1 unit. So the first blue dot: x=-2 (2 units left of origin), y = 5 (5 units up). The second blue dot: x = 4 (4 units right of origin), y=-5 (5 units down). So the slope is \(\frac{-5 - 5}{4-(-2)}=\frac{-10}{6}=-\frac{5}{3}\). Then using the point - slope form with \((-2,5)\):

\(y - 5=-\frac{5}{3}(x + 2)\)

\(y-5=-\frac{5}{3}x-\frac{10}{3}\)

\(y=-\frac{5}{3}x-\frac{10}{3}+5\)

\(y=-\frac{5}{3}x-\frac{10}{3}+\frac{15}{3}\)

\(y=-\frac{5}{3}x+\frac{5}{3}\)

Or we can write it in standard form: \(5x+3y = 5\)

Answer:

The equation of the line is \(y = -\frac{5}{3}x+\frac{5}{3}\) (or \(5x + 3y=5\))