Question

complete the following equations. enter an equation for line q: y = x

Explanation:

Step1: Identify two points on line q

From the graph, we can see that the line passes through the origin \((0,0)\) and another point, say \((2, -8)\) (by looking at the grid, when \(x = 2\), \(y=-8\)) and also \((-1,4)\) (when \(x=-1\), \(y = 4\)) and \((-2,8)\) (when \(x=-2\), \(y = 8\)). Let's use two points, for example, \((0,0)\) and \((2,-8)\).

Step2: Calculate the slope \(m\)

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Using \((x_1,y_1)=(0,0)\) and \((x_2,y_2)=(2,-8)\), we have \(m=\frac{-8 - 0}{2 - 0}=\frac{-8}{2}=- 4\). Wait, but let's check with another point. Using \((-1,4)\) and \((0,0)\), \(m=\frac{0 - 4}{0 - (-1)}=\frac{-4}{1}=-4\). So the slope \(m=-4\)? Wait, no, wait the point \((2,-8)\): when \(x = 2\), \(y=-8\), so from \((0,0)\) to \((2,-8)\), the change in \(y\) is \(-8\), change in \(x\) is \(2\), so slope is \(\frac{-8}{2}=-4\). But wait, let's check the other point \((-2,8)\): when \(x=-2\), \(y = 8\), so from \((0,0)\) to \((-2,8)\), slope is \(\frac{8-0}{-2 - 0}=\frac{8}{-2}=-4\). Wait, but maybe I made a mistake in the sign. Wait, looking at the graph, when \(x\) increases, \(y\) decreases. So the slope is negative. Wait, but let's re - examine the coordinates. Wait, the vertical axis: the top is positive or negative? Wait, the grid: the bottom part, when \(x=-2\), the \(y\) value is 8? Wait, maybe the vertical axis is reversed. Wait, maybe the \(y\) - axis: the lower part is positive? Wait, no, the standard coordinate system: \(x\) - axis horizontal (right positive), \(y\) - axis vertical (up positive). But in the graph, the point at \(x = 2\) is at \(y=-8\) (below the \(x\) - axis), and at \(x=-2\) is at \(y = 8\) (above the \(x\) - axis). So the slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take \((x_1,y_1)=(-2,8)\) and \((x_2,y_2)=(2,-8)\). Then \(m=\frac{-8 - 8}{2-(-2)}=\frac{-16}{4}=-4\). So the slope \(m=-4\)? Wait, no, wait the line passes through the origin \((0,0)\), so the equation of the line is \(y=mx\). Let's check with \(x = 2\), \(y=-8\). If \(y=mx\), then \(-8=m\times2\), so \(m=-4\). Wait, but let's check \(x=-1\), \(y = 4\): \(4=m\times(-1)\), so \(m=-4\). Yes, that works. So the equation of line \(q\) is \(y=-4x\)? Wait, no, wait maybe I mixed up the coordinates. Wait, the point labeled \(q\) is at \(x = 2\), \(y=-8\)? Wait, no, looking at the graph again, the blue dot on line \(q\) is at \(x = 2\) (horizontal axis, \(x = 2\)) and \(y=-8\) (vertical axis, \(y=-8\))? Wait, maybe the vertical axis is such that the top is negative? No, standard is up is positive. Wait, maybe the coordinates are different. Wait, let's look at the grid: each square is 1 unit. So the point at \(x = 2\) (right 2 units from origin) and \(y=-8\) (down 8 units from origin). So the slope is \(\frac{-8-0}{2 - 0}=-4\). So the equation is \(y=-4x\)? Wait, but let's check another point. When \(x=-1\), \(y = 4\) (up 4 units from origin, left 1 unit). So \(y=-4x\) gives \(y=-4\times(-1)=4\), which matches. When \(x=-2\), \(y=-4\times(-2)=8\), which also matches the point at \(x=-2\), \(y = 8\). So the slope is \(-4\), and since it passes through the origin (\(b = 0\) in \(y=mx + b\)), the equation is \(y=-4x\). Wait, but wait, maybe I made a mistake. Wait, the problem says "line q". Wait, let's re - check the points. Wait, the line passes through \((0,0)\), \((1,-4)\)? Wait, no, the blue dot is at \(x = 2\), \(y=-8\)? Wait, maybe the grid is such that each square is 1 unit, so from \((0,0)\) to \((2,-8)\), the run is 2, rise is - 8, so slope is \(-4\). So the equation is \(y=-4x\). Wait, but let's check again. If \(x = 1\), then \(y=-4\times1=-4\), but is there a point at \(x = 1\), \(y=-4\)? The graph shows a blue dot at \(x = 2\), \(y=-8\), so yes, the slope is \(-4\). So the equation of line \(q\) is \(y=-4x\). Wait, but maybe I messed up the sign. Wait, when \(x\) is positive, \(y\) is negative, when \(x\) is negative, \(y\) is positive, so the slope is negative. So the equation is \(y=-4x\).

Wait, no, wait the point at \(x=-1\) is \(y = 4\), \(x=-2\) is \(y = 8\), \(x = 2\) is \(y=-8\), \(x = 1\) is \(y=-4\). So the slope between \((-1,4)\) and \((1,-4)\) is \(\frac{-4 - 4}{1-(-1)}=\frac{-8}{2}=-4\). So the slope is \(-4\), and since it passes through the origin, the equation is \(y=-4x\).

Answer:

\(y = - 4x\)