Answer: To find the point on the line 3x + y = 8 that is closest to the point (-3,2), we need to minimize the distance between the line and the point.
Let (x, y) be the point on the line that is closest to (-3, 2). Then the vector from the point (-3, 2) to (x, y) is orthogonal (perpendicular) to the line. The direction vector of the line is <3, 1>, so the direction vector of the orthogonal vector is <-1/3, 1>.
Now we can write an equation for the line passing through (-3, 2) with the direction vector <-1/3, 1>:
(x - (-3))/(-1/3) = (y - 2)/1
Simplifying, we get:
3x + y = 11
This is the line passing through (-3, 2) that is orthogonal to the original line 3x + y = 8.
To find the intersection of these two lines, we can solve the system of equations:
3x + y = 8
3x + y = 11
Subtracting the first equation from the second, we get:
0 = 3
This is a contradiction, which means the two lines do not intersect. Therefore, the point on the line 3x + y = 8 that is closest to (-3, 2) does not exist.
However, we can still find the closest point to (-3, 2) on the line 3x + y = 8. This point will be the intersection of the line passing through (-3, 2) with the direction vector <-1/3, 1> and the line 3x + y = 8.
The equation of the line passing through (-3, 2) with the direction vector <-1/3, 1> is:
(x - (-3))/(-1/3) = (y - 2)/1
Simplifying, we get:
3x + y = 11
To find the intersection point with the line 3x + y = 8, we can solve the system of equations:
3x + y = 8
3x + y = 11
Subtracting the first equation from the second, we get:
0 = 3
This is a contradiction, which means the two lines do not intersect. Therefore, the point on the line 3x + y = 8 that is closest to (-3, 2) does not exist.
Explanation: