Find the point on the line 4x+4y+5=0 which is closest to the point (β2,β3).
Accepted Solution
A:
we know that
The point on the line 4x+4y+5=0 which is closest to the point (β2,β3)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 (β2,β3)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 (β2,β3) is the intersection of the line y=-x-1.25 and the line y=x-1