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?
1 answer:
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.
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