Pertaining to the previous question, if two of the vertices of a feasible region are at (1,3) and (6,1) how would the linear equation representing a line connecting them be found?
y=m x + b
where m=(3-1)/(1-6)=-2/5 check that.
y=-2/5 x + b now put in either point..
3=-2/5 (1)+b
b=3+2/5
so y=-2/5 x + 3+2/5
Thanks, I appreciate the help.
To find the linear equation representing a line connecting two points, you can use the slope-intercept form of a linear equation, which is y = mx + b.
First, we need to calculate the slope (m) of the line connecting the two points (1,3) and (6,1). The formula to find the slope between two points, (x1, y1) and (x2, y2), is given by:
m = (y2 - y1) / (x2 - x1)
Applying this formula to our points (1,3) and (6,1), we have:
m = (1 - 3) / (6 - 1)
= -2 / 5
Now that we have the slope (m), we can substitute it into the slope-intercept form of the linear equation along with one of the given points. Let's use the point (1,3):
y = -2/5x + b
To find the value of b (the y-intercept), we can substitute the coordinates of the point (1,3) into the equation:
3 = -2/5(1) + b
3 = -2/5 + b
Now, we can solve for b by isolating it on one side of the equation:
b = 3 + 2/5
b = 15/5 + 2/5
b = 17/5
Therefore, the linear equation representing the line connecting the points (1,3) and (6,1) is:
y = -2/5x + 17/5