SOLUTION: Determine the shortest distance from (9,5) to 5x + 3y =15

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Question 258389: Determine the shortest distance from (9,5) to 5x + 3y =15
Found 2 solutions by stanbon, richwmiller:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Determine the shortest distance from (9,5) to 5x + 3y =15
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The shortest distance is a perpendicular distance.
---
Find the slope of the given line:
y = (-5/3)x + 5
---
ANY line perpendicular to that line must
have a slope = 3/5
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Find the equation of the line with slope = 3/5 and passing thru (9,5):
5 = (3/5)9 + b
5 = (27/5) + b
b = (25/5)-(27/5)
b = -2/5
========================
Equation you want: y = (3/5)x - (2/5)
5y = 3x - 2
3x - 5y = 2
========================
Find the point of intersection of the two equations:
3x-5y = 2
5x+3y = 15
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Multiply 1st Eq. by 3; Multiply 2nd equation by 5:
9x - 15y = 6
25x- 15y = 75
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Add and solve for "x":
34x = 81
x = 81/34
--
Substitute to solve for "y":
3(81/34) - 5y = 6
243/34 - 5y = 6
5y = (243-204)/34
5y = 39/34
y = (39/170)
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The point of intersection is (81/34 , 39/170)
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Find the distance between that point and (9,5)
================================================
Cheers,
Stan H.

Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
I had to think a second about this one.
They are asking for the equation that for the line which passes through (9,5) and is perpendicular to 5x + 3y =15.
Then we need the intersection of the two lines and then the distance from the intersection to (9,5)
so find the slope of 5x + 3y =15
3y=15-5x
y=15/3-5x/3
m=-5/3
perpendicular slope would be 3/5
5=3/5(9)+b
25=27+5b
-2=5b
-2/5=b
y=3/5x-2/5
5x + 3y =15
5y=3x-2 and
3x-2=5y
3x-5y=2
5x+3y=15
9x-15y=6
25x+15y=75
34x=81
x = 81/34, y = 35/34
(9,5)
(81/34,35/34)
a^2+b^2=c^2
sqrt((9-81/34)^2 +(5-35/34)^2)=
7.7 units
Here are some other ways to do it. Some much shorter.
http://www.worsleyschool.net/science/files/linepoint/distance.html

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