# Determine a point

$$\text{ABC- triangle:} A(4,2); B(-2,1);C(3,-2)$$
Find a D point so this equality is true:

$$5\vec{AD}=2\vec{AB}-3\vec{AC}$$

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Well, you know the distance formula, I presume? – J. M. Sep 22 '11 at 14:40
(i) Compute $2\vec{AB}-3\vec{AC}$; (ii) Divide by $5$; (iii) Now you know $\vec{AD}$; (iv) Now find $D$. – André Nicolas Sep 22 '11 at 14:52
@J.M.: This problem has nothing to do with distances. – Christian Blatter Sep 22 '11 at 17:42
@Christian: Clearly my interpretation of $\vec{AD}$ as a length isn't right, then... – J. M. Sep 22 '11 at 18:06

$$\text{The given vectors } \overrightarrow{AB}=B-A\text{ and }\overrightarrow{AC}=C-A \text{ and the solution }D=A+\overrightarrow{AD}$$

Let $(x,y)$ be the coordinates of $D$. The equation

$$5\overrightarrow{AD}=2\overrightarrow{AB}-3\overrightarrow{AC}\tag{0}$$

means that

$$5(x-4,y-2)=2(-2-4,1-2)-3(3-4,-2-2),\tag{1}$$

because the vectors $\overrightarrow{AD}=D-A$, $\overrightarrow{AB}=B-A$ and $\overrightarrow{AC}=C-A$.

The vectors $5\overrightarrow{AD}=5\left( D-A\right) =\left( 5D-5A\right)$, $2\overrightarrow{AB}=\left( 2B-2A\right)$, etc.

A possible way of solving the equation $(1)$ is as follows.

$$5(x-4,y-2)=2(-2-4,1-2)-3(3-4,-2-2)$$

$$\begin{eqnarray*} &\Leftrightarrow &(5x-20,5y-10)=2(-6,-1)-3(-1,-4) \\ &\Leftrightarrow &(5x-20,5y-10)=(-12,-2)-(-3,-12) \\ &\Leftrightarrow &(5x-20,5y-10)=(-12,-2)+(3,12) \\ &\Leftrightarrow &(5x-20,5y-10)=(-12+3,-2+12) \\ &\Leftrightarrow &(5x-20,5y-10)=(-9,10) \\ &\Leftrightarrow &\left\{ \begin{array}{c} 5x-20=-9 \\ 5y-10=10 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{c} x=\frac{11}{5} \\ y=4 \end{array} \right. \end{eqnarray*}\tag{2}$$

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+1,nice explanation along with a diagram. – Quixotic Sep 22 '11 at 20:08
@FoolForMath, Thanks! – Américo Tavares Sep 22 '11 at 20:20
Nice answer, and I understood it. Thanks! :-) – Daniel Sep 23 '11 at 17:38
@Daniel: Glad to help. – Américo Tavares Sep 23 '11 at 19:12

So,let's observe picture below.first of all you will need to find point $E$...use that $E$ lies on $p(A,B)$ and that $\left\vert AB \right\vert = \left\vert BE \right\vert$. Since $p(A,C)\left\vert \right\vert p(F,E)$ we may write next equation: $\frac{y_C-y_A}{x_C-x_A}=\frac{y_E-y_F}{x_E-x_F}$ and $\left\vert EF \right\vert=3 \left\vert AC \right\vert$ so we may find point F.Since $\left\vert AF \right\vert=5 \left\vert AD \right\vert$ we may write next equations: $x_D=\frac{x_F+4x_A}{5}$ and $y_D=\frac{y_F+4y_A}{5}$

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@pedia:This problem has nothing to do with distances. – Christian Blatter Sep 22 '11 at 17:42

Recall that the vector $\overrightarrow{PQ}$ is the difference of two points $Q{-}P$. In this way, $$5\overrightarrow{AD}=2\overrightarrow{AB}-3\overrightarrow{AC}$$ becomes $$5(D-A)=2(B-A)-3(C-A)$$ All that is left is to solve for $D$.

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