Answer:
A- The point-slope form of a line parallel to another line is given by y - y1 = m(x - x1), where m is the slope of the original line. The slope of the line x+2y=6 is -1/2. So, the equation of the line parallel to x+2y=6 through (8,3) is y - 3 = -1/2 (x - 8), which can be written in standard form as -2x + y = -11.
B- The point-slope form of a line perpendicular to another line is given by y - y1 = -1/m(x - x1), where m is the slope of the original line. The slope of the line x+2y=6 is -1/2. So, the equation of the line perpendicular to x+2y=6 through (8,3) is y - 3 = 1/2 (x - 8), which can be written in standard form as 2x + y = 15.
C- To graph the three lines, we can plot the x- and y-intercepts of each line and then draw a line through the points.
The x-intercepts are (-11/2,0), (0,-15), (0,-11) and the y-intercepts are (0,3), (0,7.5) and (0,5.5) respectively.
For a more detailed graph you can use a graphing software or calculator.