Question

trazar la recta.
$y = -\frac{2}{3}x + 1$

Explanation:

Step1: Identify the y-intercept

The equation is in slope - intercept form \(y = mx + b\), where \(b\) is the y - intercept. For \(y=-\frac{2}{3}x + 1\), when \(x = 0\), \(y=1\). So one point on the line is \((0,1)\).

Step2: Use the slope to find another point

The slope \(m=-\frac{2}{3}\), which means for a run of \(3\) (change in \(x\)) to the right, the rise is \(- 2\) (change in \(y\)). Starting from \((0,1)\), if we move \(x = 0+3 = 3\), then \(y=1+( - 2)=-1\). So another point is \((3,-1)\). We can also use a run of \(- 3\) (move 3 units to the left) and a rise of \(2\) (move 2 units up). Starting from \((0,1)\), if \(x = 0-3=-3\), then \(y = 1 + 2=3\), so the point \((-3,3)\) is also on the line.

Step3: Plot the points and draw the line

Plot the points \((0,1)\), \((3,-1)\) (or \((-3,3)\)) on the coordinate plane and then draw a straight line passing through these points.

(Note: To actually draw the line on the given grid, mark the points \((0,1)\), \((3, - 1)\) or other points found using the slope - intercept form and connect them with a straight edge.)

Answer:

To draw the line \(y =-\frac{2}{3}x + 1\), plot the y - intercept \((0,1)\) and use the slope \(-\frac{2}{3}\) to find another point (e.g., \((3,-1)\) or \((-3,3)\)) and then draw a straight line through these points.