Q:

Find the point on the line 4x+4y+5=0 which is closest to the point (βˆ’2,βˆ’3).

Accepted Solution

A:
we know that

The point on the line 4x+4y+5=0 which is closest to the point (βˆ’2,βˆ’3)Β will be located on the line perpendicular to the given line and will go through the given point

step 1
the slopes of two perpendicular lines
are
m1*m2=-1
given line
4x+4y+5=0------> 4y=-4x-5----> y=-x-1.25
m1=-1
so
m2=1

step 2
find the equation of the line with m2=1 and point (βˆ’2,βˆ’3)Β 
y-y1=m*(x-x1)-----> y+3=(1)*(x+2)----> y=x+2-3-----> y=x-1

step 3Β 
The point on the line 4x+4y+5=0 which is closest to the point (βˆ’2,βˆ’3) is the intersection of the line y=-x-1.25 and the line y=x-1

using a graph tool
see the attached figure

the solution is the point (-0.125,-1.125)