What am I doing wrong?
What you mean what did you do
Point G is located at (3, −1), and point H is located at (−2, 3). Find the point that is 2/3 the distance from point G to point H. |dw:1419870489375:dw| \(\bf G(3,-1)\qquad H(-2,3)\\ \quad \\ \quad \\ \color{blue}{\cfrac{HP}{GP}=\cfrac{ratio1}{ratio2}\implies ratio2\cdot HP=ratio1\cdot GP\quad \textit{dividing by P}\\ \quad \\ ratio2\cdot H=ratio1\cdot G\implies} 2(-2,3)=3(3,-1)\\ \quad \\\qquad \color{blue}{P=\left(\cfrac{\textit{sum of "x" values}}{ratio1+ratio2}\quad ,\quad \cfrac{\textit{sum of "y" values}}{ratio1+ratio2}\right)}\\ \quad \\ \qquad thus\qquad \\ \quad \\ P=\left(\cfrac{(2\cdot -2)+(3\cdot 3)}{2+3}\quad ,\quad \cfrac{(2\cdot 3)+(3\cdot -1)}{2+3}\right)\) I get \(\bf (1, 0.6)\)..
Hold on..let me get the picture..
However, (1, 0.6) is incorrect..
\(\Large \dfrac{3\times 2 + (-2)\times 3}{2+3}\), \(\Large \dfrac{(-1)\times 2 + 3\times 3}{2+3}\) ^^ that should be your setup
for P
Oh..
I get (3.6, 0.6).
I wish I could help but I got the open study app on my phone and it sucks lol I can't see the if that's a chart or if it's like what I see
\(\Large \dfrac{mx_1 +nx_2}{m+n}\), \(\Large \dfrac{my_1 +ny_2}{m+n}\) for n:m ratio thats the general form
Oops, I didn't post the whole question.. Point G is located at (3, −1), and point H is located at (−2, 3). Find the point that is 2/3 the distance from point G to point H.
That's (3, -1) and (-2, 3).
\(m = 2\) \(n = 3\) \(x_1 = 3\) \(y_1 = -1\) \(x_2 = -2\) \(y_2 = 3\) Correct..?
'2/3 the distance from point G to point H' m:n = 2:3 yes all those are correct
Which gives me: \((\dfrac{m x_1 + n x_2}{m + n}, \dfrac{m y_1 + n y_2}{m + n})\) \((\dfrac{(2)(3) + (3)(-2)}{2 + 3}, \dfrac{(2)(-1) + (3)(3)}{2 + 3})\) \((\dfrac{6 + -6}{5}, \dfrac{-2 + 9}{5})\) \((\dfrac{0}{5}, \dfrac{7}{5})\) \((0, 1.4)\) That answer is still wrong.. @hartnn
why do i get 0,8/5 ?
nah, 0,7/5 is correct
That's not one of my options..
(0.33, -1.67) (-0.33, 1.67) (-0.5, -1) (6.33, -2.67)
I think (−0.33, 1.67) is my answer..
oh wait!
2/3 the distance so, 2/3 : 1/3 = 2:1 :3
m:n = 2:1
now try again :P
Oh..
\((\dfrac{m x_1 + n x_2}{m + n}, \dfrac{m y_1 + n y_2}{m + n})\) \((\dfrac{(2)(3) + (1)(-2)}{2 + 1}, \dfrac{(2)(-1) + (1)(3)}{2 + 1})\) \((\dfrac{6 + (-2)}{3}, \dfrac{-2 + 3}{3})\) \((\dfrac{-12}{3}, \dfrac{1}{3})\) \((-4, 0. \overline{33})\)
I'm still doing something wrong..
yeah, you swapped m and n
But I know the answer is (-0.33, 1.67)..because if you graph all the points with the two points..then (-0.33, 1.67) is the only one that lies in the middle of the two points..
|dw:1419871767756:dw|
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