91Ó°ÊÓ

To find the slope of the line and a point on it, we first need to convert the given parametric equations into a standard linear equation of the form $$y = mx + b$$. First, solve for $$t$$ from either equation and substitute that into the other equation. From the \(x\) equation: $$x = 3 + t \Rightarrow t = x - 3$$ Now, substitute this into the $$y$$ equation: $$y = 1 - (x - 3)$$

Simplify the equation by distributing the subtraction and combining like terms: $$y = 1 - x + 3$$ $$y = -x + 4$$ We now have the standard linear equation: $$y = -x + 4$$

Given the equation $$y = -x + 4$$, we can see the slope of the line (the coefficient of $$x$$) is equal to $$m = -1$$

Now we need to find a point on the line by choosing a value for $$t$$ in the parametric equations. Let $$t = 0$$, then $$x = 3 + 0 = 3$$ $$y = 1 - 0 = 1$$ Thus, the point $$(3, 1)$$ lies on the line.

To graph the line, we can begin by plotting the point $$(3, 1)$$ on the coordinate plane. Then, we can use the slope, which is $$-1$$, to find additional points on the line. Since the slope is a ratio of the vertical change ($$\Delta y$$) to the horizontal change ($$\Delta x$$), a slope of $$-1$$ means that for each unit we move to the right, we move down one unit. Thus, we can find other points on the line by adding/subtracting its slope to one point giving us new points such as $$(4, 0)$$, $$(2, 2)$$, and many others. Finally, we can draw the line that passes through these points. The slope of the line is $$m = -1$$, with a point on the line being $$(3, 1)$$. The graph of the line can be drawn by plotting these points and following the slope of $$-1$$.