Avery can run at 10 uph. the bank of a river is represented by the line 4x + 3y = 12, and avery is at (7, 5). how much time does avery need to reach the river?

The minimum distance will be along a perpendicular line to the river that passes through the point (7,5)

4x+3y=12

3y=-4x+12

y=-4x/3+12/3

So a line perpendicular to the bank will be:

y=3x/4+b, and we need it to pass through (7,5) so

5=3(7)/4+b

5=21/4+b

20/4-21/4=b

-1/4=b so the perpendicular line is:

y=3x/4-1/4

So now we want to know the point where this perpendicular line meets with the river bank.  When it does y=y so we can say:

(3x-1)/4=(-4x+12)/3  cross multiply

3(3x-1)=4(-4x+12)

9x-3=-16x+48

25x=51

x=51/25

x=2.04

y=(3x-1)/4

y=(3*2.04-1)/4

y=1.28

So now that we know the point on the river that is closest to Avery we can calculate his distance from that point...

d^2=(x2-x1)^2+(y2-y1)^2

d^2=(7-2.04)^2+(5-1.28)^2

d^2=38.44

d=√38.44

d=6.2 units

Since he can run at 10 uph...

t=d/v

t=6.2/10

t=0.62 hours  (37 min 12 sec)

So it will take him 0.62 hours or 37 minutes and 12 seconds for him to reach the river.

Ithink it's b sorry if i'm wrong