SOLUTION: Calculate the shortest distance from the given line and point povided. 2x-y=3 (2,3) I subtracted 2 from 2x-y=3 and got -y=-2+3, I made them postitive and got y=1 I plu

Algebra ->  Graphs -> SOLUTION: Calculate the shortest distance from the given line and point povided. 2x-y=3 (2,3) I subtracted 2 from 2x-y=3 and got -y=-2+3, I made them postitive and got y=1 I plu      Log On


   

Question 179956: Calculate the shortest distance from the given line and point povided.
2x-y=3 (2,3)

I subtracted 2 from 2x-y=3 and got -y=-2+3, I made them postitive and got y=1
I plugged 2 in for x then got seven, I put 2(1)-7=3 and recieved 5. I was told this is incorrect but i dont know any other way to do it...any help would be helpful.
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
find the equation of the line that passes thru (2,3) and is perpendicular to the given line
:
the slope of the perpendicular line will have a negative reciprical to the given line. 2x-y=3 --->y=2x-3 so m= for this line is 2. the slope for the perpendicular line is -1/2
:
y-3=(-1/2)(x-2)
:
y-3=(-1/2)x+1--->y=(-1/2)x+4--->x+2y=8
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now we have to find the intersection of these two lines
:
2x-y=3....eq 1
x+2y=8...eq 2
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multiply eq 1 by 2 and add all terms of the equations together.
:
4x-2y=6....eq 1 revised
x+2y=8....eq 2
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y terms are eliminated becaue -2y+2y=0. We are left with 4x+x=14
:
5x=14
:
x=14/5
:
plug this value back into eq 1 or 2. I chose eq 2
:
(14/5)+2y=8
:
2y=26/5:
y=26/10=13/5
:
Now we can use the distance formula: we have point (2,3) and point (14/5,13/5)
:
d=sqrt%28%2814%2F5-2%29%5E2%2B%2813%2F5-3%29%5E2%29
:
d=sqrt%28%284%2F5%29%5E2%2B%283%2F5%29%5E2%29
:
d=sqrt%28%2816%2F25%29%2B%289%2F25%29%29
:
d=sqrt%281%29=1
:
distance between point (2,3) and line 2x-y=3 is 1
:
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C2x-3%2C%28-1%2F2%29x%2B4%29