PLEASE HELP, WILL FAN AND GIVE MEDAL!!!!!! Point G is located at (3, -1) and point H is located at (-2, 3). Find the point that is the distance from point G to point H.
@sourwing
"Find the point that is the distance from" ^ is there anything missing there?
or you just need how long from G to H?
\(\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ G&({\color{red}{ 3}}\quad ,&{\color{blue}{ -1}})\quad H&({\color{red}{ -2}}\quad ,&{\color{blue}{ 3}}) \end{array}\qquad d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}\)
oh its 2/3 the distance, sorry bout that
@jdoe0001
hmm 2/3 the distance? from G to H? any ideas what that means?
well we have to find the point where that would be
|dw:1402179798046:dw| would that be about right?
i think so but i need the coordinates
would it be fair to say the point P is at a ratio of 2:1 from G to H?
if from say G to P, is 2/3 of the whole line and from P to H is 1/3 only of the whole line then the distance GP will be at a ratio of 2:1, that is, twice as much as the distance PH .... is that fair to say?
well its asking 2/3 of the distance from G to P so the point would lie on line GP
@mathslover
@vishweshshrimali5
@mathmate
Hi @kd7023s5
hello..
Okay lets see what do we have ..... okay lets assume the required point be P (x,y). Okay ?
ok
wait.. i think it was a mistake on my part because @jdoe0001 was right but i just needed the coordinates of point p
sorry...
its ok
Answer me this would this point P necessarily lie on line GP ?
yeah
No, its not necessary. See: |dw:1402238160153:dw| (the circle is drawn in such a way that its centre is H and HP is its radius).
it doesnt look like h is the center..
my answer was (-0.33, 1.67)
@kd7023s5 please forget my replies. I took the question to a little higher level than required. It would be best if @jdoe0001 helped you out. Sorry
Its just that if you would follow my solution you will get infinitely many such points P.
youre right but the question is asking for a point on the line
and its ok, i was right
Yeah I have verified it. Good work
Did we arrive at the answer?
yeah
okay
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