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
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-> 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
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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) (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
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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
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y-3=(-1/2)(x-2)
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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
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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.
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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
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5x=14
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x=14/5
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plug this value back into eq 1 or 2. I chose eq 2
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(14/5)+2y=8
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2y=26/5:
y=26/10=13/5
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Now we can use the distance formula: we have point (2,3) and point (14/5,13/5)
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distance between point (2,3) and line 2x-y=3 is
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