*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 \(\displaystyle \lambda = -\frac{5}{9}\)