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The Distance between the point (5,4) and the Xaxis is ...... Length unit
~~~~~~~~
Guys I need an Urgent Help xD ..
 one year ago
 one year ago
The Distance between the point (5,4) and the Xaxis is ...... Length unit ~~~~~~~~ Guys I need an Urgent Help xD ..
 one year ago
 one year ago

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ZeHanzBest ResponseYou've already chosen the best response.1
Make a drawing. Put the point (5,40) in its proper position and then ask yourself: what is the shortest posible way to get from there to the xaxis?
 one year ago

EyadBest ResponseYou've already chosen the best response.0
@ZeHanz : :D that answer was the shortest and Hintless answer ?! I think i'am not able to measure it I should use da law ...
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.1
In strict mathematical terms (and offered here this way so as to show the concept most clearly and completely) is to find the perpendicular distance from point (5, 4) to the line y = 0. Therefore, you would take as the measure of that slope, the negative reciprocal of the slope of line y = 0, and that is an "undefined" slope, so we are talking about a vertical line or a vertical distance. The problem becomes much, much easier at this point because we don't have to come up with the equation of a line now. We can just find the distance from (5, 4) and (5, 0), and that distance is merely 4 or 4.
 one year ago

EyadBest ResponseYou've already chosen the best response.0
@tcarroll010 ,Well That was ma answer (thats ma sis's question) However her question has the last solution which is the distance will be 10 ! Maybe we should use her lesson concept which is using da law : D= Root(x2x1)+(y2_y1) ..
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.1
Where "y = 0" comes from is that that is the equation for the xaxis. You are essentially asking how to find the distance from a point to a line, so that is always a perpendicular distance. You can then generalize that concept to find the distance of any point to any line.
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
I thought I didn't give you a "hintless" answer ;) Also there is no need to use any kind of "law". It's just counting the boxes. (5, 4) means you are 4 units below the xaxis, so that is the answer. It would be rather embarrassing to "calculate" it with the distance formula IMO.
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.1
In general, you would use the distance formula for the distance o a point to a line. But here, the xcoordinates will be the same, so the distance is just the "y difference". For this specific problem. That is your quickest and legitimate approach. No need to overcomplicate. I mentioned the the method in the first post as the general method. here, just take the "y difference".
 one year ago

EyadBest ResponseYou've already chosen the best response.0
I know ,But da final answer should be 10 .!
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.1
You could always use the disatance formula. You would find that: d = [(5  5)^2 + (4  0)^2]^(1/2) and that will come down to: (4^2)^(1/2) = 4 That's the hard way, but that will always work. Here, we can just take the difference of the "y values"
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
It's like saying: 2+2=4, but it should be 10.
 one year ago

EyadBest ResponseYou've already chosen the best response.0
Alright ,Maybe Da final answer is misprinted ..
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.1
and since the point on the xaxis that you are measring to is (5, 0), all you are really doing is measuring the distance from 4 to 0. Which is merely 4.
 one year ago

tcarroll010Best ResponseYou've already chosen the best response.1
thx for the recognition and good luck in all of your studies!
 one year ago
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