Isosceles triangle

[IloveManUtd]

An isosceles triangle POT is such that PO=PT, O is the origin, T is the point (t,0) and P lies on the line y=mx, where m is a positive value. If the point Q is taken so that OTPQ forms a parallelogram, find the value of m when OQ = 2 0T.

I've already found the coordinates of P to be (t/2 , mt/2) and Q to be (3t/2 , mt/2). Please help me to find the value of m. Thanks

 

[soroban]

Hello, IloveManUtd!

$\displaystyle \text{An isosceles triangle }POT\text{ is such that }PO=PT,\;O\text{ is the origin, }\:T\text{ is the point }(t,0)$

. . $\displaystyle \text{and }P\text{ lies on the line: }\:y\:=\:mx\:\text{ where }m > 0.$

$\displaystyle \text{If the point }Q\text{ is taken so that }OTPQ \text{ forms a parallelogram,}$

. . $\displaystyle \text{find the value of }m\text{ when }\:OQ \,=\, 2\cdot 0T$


$\displaystyle \text{I've already found the coordinates of }P\text{ to be }\,\left(\frac{t}{2},\;\frac{mt}{2}\right)$ . Yes!
. . $\displaystyle \text{and }Q \text{ to be: }\:\left(\frac{3t}{2},\:\frac{mt}{2}\right)$ . No


$\displaystyle \text{Please help me to find the value of }m.$

We have this isosceles triangle:

            |       /
            |     P/
            |     *
            |    / \
            |   /   \
            |  /     \
            | /       \
            |/         \
        - - * - - - - - * - -
           /O     t     T
          / |

$\displaystyle \text{We are given parallelogram }OTPQ$

$\displaystyle \text{The vertices are always given "in order",}$
. . $\displaystyle \text{so we know where vertex }Q\text{ must be.}$

      Q     |     P
      * - - | - - *
       \    |      \
        \   |       \
         \  |        \
          \ |         \
           \|          \
      - - - * - - - - - * - -
            O     t     T

\(\displaystyle \text{Hence, }Q\text{ is }t\text{ units to the left of }P \;\hdots\;Q\left(-\frac{t}{2},\:\frac{mt}{2}\right)\)


$\displaystyle \text{We want: }\:OQ \:=\:2\cdot OT$


$\displaystyle \text{Hence: }\:\sqrt{\left(-\frac{t}{2}\right)^2 + \left(\frac{mt}{2}\right)^2} \:=\:2t \quad\Rightarrow\quad \frac{t^2}{4} + \frac{m^2t^2}{4} \:=\:4t^2$


\(\displaystyle \text{Multiply by }\frac{4}{t^2}:\;\;1 + m^2 \:=\:16 \quad\Rightarfrow\quad m^2 \:=\:15 \quad\Rightarrow\quad \boxed{m \:=\:\sqrt{15}}\)