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Mathematics 15 Online
OpenStudy (anonymous):

In the coordinate plane, three vertices of a rectangle PQRS are P(0,0), Q(0,b), and S(c,0). what are the coordinates of point R

OpenStudy (anonymous):

You mean "rectangle" rather than "triangle", don't you ?

OpenStudy (anonymous):

Yeah I meant to say rectangle

OpenStudy (amistre64):

make a plot of it q r p s q0 b rx y what corresponds to x and y ? p0 0 sc 0

OpenStudy (anonymous):

Since PQRS is a rectangle then : \[\overrightarrow{SR}=\overrightarrow{PQ}\] So : \[\begin{pmatrix}x_R-x_S\\y_R-y_S\end{pmatrix}=\begin{pmatrix}x_Q-x_P\\y_Q-y_P\end{pmatrix}\] So : \[\begin{pmatrix}x_R-c\\y_R-0\end{pmatrix}=\begin{pmatrix}b-0\\0-0\end{pmatrix}\] So : \[x_R=b+c \\y_R=0 \]

OpenStudy (amistre64):

your calculations appear to be off, or im misreading your end results

OpenStudy (amistre64):

R: (b+c, 0) ??

OpenStudy (anonymous):

I got (c,b) as the answer

OpenStudy (amistre64):

so did i :)

OpenStudy (anonymous):

I have done a mistake, but the method it what is more important :

OpenStudy (amistre64):

if we consider these are parallel lines, the vector stuff would turn out as: from P to Q applied at P x = 0 + t(0) y = 0 + t(b) from S to R is the same vector applied at S xr = c + t(0) yr = 0 + t(b) by letting t represent the length of 1 vector, a unit length: t = 1 xr = c + 0 yr = 0 + b

OpenStudy (anonymous):

thankyou

OpenStudy (amistre64):

youre welcome, if you dont need to run vectors for this, id just go with the basic general plot construction

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