The distance of point (8,5) from the straight line 3x+4y+1=0 is equal a) 7b) 9c)10d)8

9

Step-by-step 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

there is no question content here, but you for the free points! : d