SOLUTION: Find the distance from (6,-3)to the line defined by y= 2x-5. Express as a radical or a number rounded to the nearest hundreth

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Question 849870: Find the distance from (6,-3)to the line defined by y= 2x-5. Express as a radical or a number rounded to the nearest hundreth
Found 3 solutions by Alan3354, Fombitz, Edwin McCravy:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Step 1, find the line perpendicular to y = 2x-5 thru (6,3), point A.
Step 2, find the intersection of the 2 lines, point B.
Step 3, find the distance AB.
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Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
Do you mean the minimum distance to the line?
If so, the minimum distance is a perpendicular line going through the point.
A perpendicular line would have the form,

Use the point to solve for .



.
.
.
Find the intersection of the two lines.


Set them equal to each other.




Then from either equation, solve for .

Now find the distance from (6,-3) to (2,-1) using the distance formula,
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
They did it the long way.  

The formula for the distance from a point to a line is



where the equation of the line is Ax+By+C=0 and the point is (x1,y1)

The point is (x1,y1) = (6,-3)
The line's equation is y=2x-5. we must get that in the form
Ax+By+C=0

      y=2x-5.  We subtract the entire right side from both sides
 
-2x+y+5=0

A=-2, B=1, C=5, (x1,y1) = (6,-3)











We can rationalize the denominator by multiplying by .

.

That's about 4.47, so let's draw the graph as a check:



That green line looks like it's about 4.47 units in length.

Edwin
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