Question

draw a line through (2,2) with a slope of $\frac{2}{3}$. mark at least three additional points on the line.

Explanation:

Step1: Recall slope - point form

The slope - point form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope. Here $x_1 = 2,y_1=2,m=\frac{2}{3}$. So the equation of the line is $y - 2=\frac{2}{3}(x - 2)$.

Step2: Find additional points

• For $x=5$:

Substitute $x = 5$ into $y - 2=\frac{2}{3}(x - 2)$. Then $y-2=\frac{2}{3}(5 - 2)=\frac{2}{3}\times3 = 2$, so $y=4$. The point is $(5,4)$.

• For $x=-1$:

Substitute $x=-1$ into $y - 2=\frac{2}{3}(x - 2)$. Then $y - 2=\frac{2}{3}(-1 - 2)=\frac{2}{3}\times(-3)=-2$, so $y = 0$. The point is $(-1,0)$.

• For $x = 8$:

Substitute $x = 8$ into $y - 2=\frac{2}{3}(x - 2)$. Then $y-2=\frac{2}{3}(8 - 2)=\frac{2}{3}\times6 = 4$, so $y=6$. The point is $(8,6)$.

To draw the line, first plot the point $(2,2)$ on the coordinate - plane. Then plot the additional points $(-1,0),(5,4),(8,6)$ and draw a straight line passing through all these points.

Answer:

The additional points can be $(-1,0),(5,4),(8,6)$ (answers may vary). Plot these points along with $(2,2)$ and draw a straight line through them.