Answer:
9
Explanation:
So we need to find the point on 3x+4y+1=0 such that when connecting that point to (8,5) the lines that interest are perpendicular ones.
First step solve 3x+4y+1=0 for y.
subract 3x and 1 on both sides: 4y=-3x-1
divide both sides by 4: y=-3/4x-1/4
So a line that is perpendicular to this one is 4/3
So we have the perpendicular line is in the form of y=4/3 x+ b
now we do want this to go through (8,5)
5=4/3 (8)+b
5=32/3+b
5-32/3=b
(15-32)/3=b
-17/3=b
So the perpendicular line we are looking at is y=4/3 x -17/3 .
Now I can find the point I talked about in my first sentence if I find the intersection of the line I just got and the line we started with. I'm going to just sub one into the other since they are both solve for y now.
-3/4 x-1/4=4/3 x-17/3
add 1/4 on both sides
-3/4 x =4/3 x-65/12
subtract 4/3 x on both sides
-3/4 x-4/3 x=-65/12
simplify
-25/12 x =-65/12
25x=65
x=65/25
x=13/5
Now find y by plugging this into y=-3/4 x-1/4 giving you y=-11/5
So we want to actually just find the distance between (13/5,-11/5) and (8,5)
which is sqrt((27/5)^2+(36/5)^2)=9