Question

cl 2 - 121. graph the segment that connects the points a(-4, 8) and b(6, 3). a. what is the slope of (overline{ab})? b. write an equation for the line that connects points a and b. c. write an equation for a line that is parallel to (overline{ab}). d. write an equation for a line that is perpendicular to (overline{ab}).

Explanation:

Step1: Calculate the slope of line segment AB

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. Given $A(-4,8)$ and $B(6,3)$, we have $x_1=-4,y_1 = 8,x_2=6,y_2 = 3$. Then $m=\frac{3 - 8}{6-(-4)}=\frac{-5}{10}=-\frac{1}{2}$.

Step2: Find the equation of the line connecting A and B

The point - slope form of a line is $y - y_1=m(x - x_1)$. Using point $A(-4,8)$ and $m =-\frac{1}{2}$, we get $y - 8=-\frac{1}{2}(x + 4)$. Expanding gives $y-8=-\frac{1}{2}x-2$, and then $y=-\frac{1}{2}x + 6$.

Step3: Find the equation of a line parallel to AB

Parallel lines have the same slope. Let the equation of the parallel line be $y=-\frac{1}{2}x + b$. Without a given point on this parallel line, we leave it in this general form.

Step4: Find the equation of a line perpendicular to AB

The slope of a line perpendicular to a line with slope $m$ is the negative reciprocal. The slope of the perpendicular line is $m'=2$. Using the point - slope form $y - y_1=m'(x - x_1)$ (again, without a given point, we leave it as $y = 2x + c$ in general form).

Answer:

a. The slope of $\overline{AB}$ is $-\frac{1}{2}$.
b. The equation of the line that connects points A and B is $y=-\frac{1}{2}x + 6$.
c. An equation for a line that is parallel to $\overline{AB}$ is $y=-\frac{1}{2}x + b$ (where $b$ is a real - number constant).
d. An equation for a line that is perpendicular to $\overline{AB}$ is $y = 2x + c$ (where $c$ is a real - number constant).