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[SOLVED] Three of the coordinates of the parallelogram STUV

karush

Well-known member
Jan 31, 2012
2,718
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
Jan 26, 2012
1,403
You have [tex]\displaystyle \begin{align*} x = -4 - 9\lambda \end{align*}[/tex] and [tex]\displaystyle \begin{align*} y = 6 - 9\lambda \end{align*}[/tex], so surely if you have the point [tex]\displaystyle \begin{align*} (x, y) = (1, 11) \end{align*}[/tex] you can find [tex]\displaystyle \begin{align*} \lambda \end{align*}[/tex]...
 

eddybob123

Active member
Aug 18, 2013
76
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)
 
Last edited:

soroban

Well-known member
Feb 2, 2012
409
Hello, karush!

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

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

The sketch locates point [tex]V.[/tex]
Code:
                  |       (5,15)
                  |       U o
                  |       .   *
                  |     .       * (7.7)
                  |   .           o T
                  | .           * :
                  .           *   :
                . |         *     :
              .   |       *       :
            .     |     *         :-9
          .       |   *           :
      V o         | *             :
    ------.-------*---------------:----
            .   * |               :
            S o - | - - - - - - - *
           (-2,-2)|    -9
Going from [tex]T[/tex] to [tex]S[/tex], we move down 9 and left 9.

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

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Hello, karush!


The sketch locates point [tex]V.[/tex]
Code:
                  |       (5,15)
                  |       U o
                  |       .   *
                  |     .       * (7.7)
                  |   .           o T
                  | .           * :
                  .           *   :
                . |         *     :
              .   |       *       :
            .     |     *         :-9
          .       |   *           :
      V o         | *             :
    ------.-------*---------------:----
            .   * |               :
            S o - | - - - - - - - *
           (-2,-2)|    -9
Going from [tex]T[/tex] to [tex]S[/tex], we move down 9 and left 9.

Doing the same from [tex]U[/tex] we arrive at [tex]V(-4,6).[/tex]
As impressive as your coding skills are, I have to ask, didn't the OP already do all of this?
 

karush

Well-known member
Jan 31, 2012
2,718
You have [tex]\displaystyle \begin{align*} x = -4 - 9\lambda \end{align*}[/tex] and [tex]\displaystyle \begin{align*} y = 6 - 9\lambda \end{align*}[/tex], so surely if you have the point [tex]\displaystyle \begin{align*} (x, y) = (1, 11) \end{align*}[/tex] you can find [tex]\displaystyle \begin{align*} \lambda \end{align*}[/tex]...
OK from this I get \(\displaystyle \lambda = -\frac{5}{9}\)