Quadrants and Probability

We can find the probability that a randomly chosen point from this segment is in Quadrant four by dividing the length of the smaller segment contained within Quadrant IV by the length of the total segment. Using the distance formula, we find the length of the segment:

$d=\sqrt{(6-(-2))^2+(-3-1)^2}=\sqrt{8^2+(-4)^2}=\sqrt{64+16}=\sqrt{80}=4\sqrt{5}$

Next, we need to determine where the line segment crosses the $x-axes$ and the $y-axes$ . One way to do this is to write an equation through the endpoints.

The slope of the line is $\frac{1-(-3)}{6-(-2)}=\frac{4}{8}=\frac{1}{2}$ . Then for an equation of the form $y=mx+b$ , we can substitute $\frac{1}{2}$ for $m$ , and $(6,1) for (x,y)$ to find $b$ which is the y-intercept.

$1=\frac{1}{2}(6)+b \implies b=-2$

Thus, an equation for the line containing the segment is $y=\frac{1}{2}x-2$ , and the y-intercept is $-2$ . To find the x-intercept, we substitute $0$ for $y$ in the equation and solve for $x$ :

$0=\frac{1}{2}x-2 \implies x=4$

Thus, the line segment with endpoints $(0,2) and (4,0)$ is the part of the given segment that is in Quadrant IV. Using the distance formula, we find the length of the segment in Quadrant IV:

$\sqrt{(4-0)^2+(0-(-2))^2}=\sqrt{4^2+2^2}=2\sqrt{5}$

The probability that a point along the line segment is in Quadrant IV is $\boxed{\frac{2\sqrt{5}}{4\sqrt{5}}=50\%}$