# [SOLVED]Three of the coordinates of the parallelogram STUV

#### karush

##### Well-known member
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

##### Well-known member
MHB Math Helper
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

##### Active member
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:
$$\theta_{ab}=\arccos\frac{a\cdot b}{\mid \mid a\mid \mid \mid \mid b\mid \mid }$$

Last edited:

#### soroban

##### Well-known member
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 $$V.$$
Code:
                  |       (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

##### Well-known member
MHB Math Helper
Hello, karush!

The sketch locates point $$V.$$
Code:
                  |       (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

##### Well-known member
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 $$\displaystyle \lambda = -\frac{5}{9}$$