*3 coordinates of parallelogram STUV

[karush]

View attachment 1233

(a) $\vec{ST} = \pmatrix{9 \\ 9}$
so $V=(5,15)-(9,9)=(-4,6)$

(b) $UV = \pmatrix{-4,6}-\lambda \pmatrix{9,9}$

(c) eq of line $UV$ is $y=x+10$ so from position vector
$\pmatrix{1 \\11}$ we have $11=1+10$

didn't know how to find the value of $\lambda$

(d) ?

 

[Prove It]

You have \displaystyle \begin{align*} x = -4 - 9\lambda \end{align*} and \displaystyle \begin{align*} y = 6 - 9\lambda \end{align*}, so surely if you have the point \displaystyle \begin{align*} (x, y) = (1, 11) \end{align*} you can find \displaystyle \begin{align*} \lambda \end{align*}...

 

[eddybob123]

For d) i), you can easily apply the distance formula in elementary geometry. You should get
$$\sqrt{(a-1)^2+(17-11)^2}=2\sqrt{13}$$
From that point it's just algebra.
For ii), use this formula involving vector dot products:
$$\theta_{ab}=\arccos\frac{a\cdot b}{\mid \mid a\mid \mid \mid \mid b\mid \mid }$$

(Bandit)

 

[soroban]

Hello, karush!

\text{20. Three of the coordinates of parallelogram }STUV
. . . .\text{are: }\:S(\text{-}2,\text{-}2),\:T(7,7),\:U(5,15)

\text{(a) Find the vector }\vec{ST}\text{ and hence the coordinates of }V.

The sketch locates point V.

                  |       (5,15)
                  |       U o
                  |       .   *
                  |     .       * (7.7)
                  |   .           o T
                  | .           * :
                  .           *   :
                . |         *     :
              .   |       *       :
            .     |     *         :-9
          .       |   *           :
      V o         | *             :
    ------.-------*---------------:----
            .   * |               :
            S o - | - - - - - - - *
           (-2,-2)|    -9

Going from T to S, we move down 9 and left 9.

Doing the same from U we arrive at V(-4,6).

 

[Prove It]

Hello, karush!


The sketch locates point V.

                  |       (5,15)
                  |       U o
                  |       .   *
                  |     .       * (7.7)
                  |   .           o T
                  | .           * :
                  .           *   :
                . |         *     :
              .   |       *       :
            .     |     *         :-9
          .       |   *           :
      V o         | *             :
    ------.-------*---------------:----
            .   * |               :
            S o - | - - - - - - - *
           (-2,-2)|    -9

Going from T to S, we move down 9 and left 9.

Doing the same from U we arrive at V(-4,6).

As impressive as your coding skills are, I have to ask, didn't the OP already do all of this?

 

[karush]

You have \displaystyle \begin{align*} x = -4 - 9\lambda \end{align*} and \displaystyle \begin{align*} y = 6 - 9\lambda \end{align*}, so surely if you have the point \displaystyle \begin{align*} (x, y) = (1, 11) \end{align*} you can find \displaystyle \begin{align*} \lambda \end{align*}...

OK from this I get $$\lambda = -\frac{5}{9}$$