|
Question 62752: Hi, this geometry problem was given to me as a homework assignment and I'm really having a difficult time figuring it out. I'm not doing so well in Geometry right now, and I could really use the extra help.
Find the distance from (7,4) to y = -x + 5. Your answer should be an actual number, not a line.
Thank you so much!
Answer by uma(370)   (Show Source):
You can put this solution on YOUR website! The distance from a point (x1,y1) to a line ax + by + c = 0 is given by
(ax1 + by1 + c)/sqrt(a^2 + b^2)
Here the point is (7,4).
The line is y = -x + 5
==> y + x - 5 = -x + 5 + x - 5 [adding x - 5 to both the sides]
==> x + y - 5 = 0
Here a = 1, b = 1 and c = -5.
Thus the distance between the given point and the given line is
(7*1 + 4*1 - 5)/sqrt (1^2 + 1^2)
= (7 + 4 - 5)/sqrt(1 + 1)
= 6/sqrt(2)
multiplying by sqrt 2 in the numerator and the denominator we get,
= 6*sqrt(2)/sqrt(2)*sqrt(2)
= 6*sqrt(2)/2
= 3 sqrt(2).
Thus the distance = 3sqrt(2) units
Good Luck!!!
|
|
|
| |
