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# Help Please

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Let us have four distinct collinear points A, B, C and D on the Cartesian plane. The point C is such that AB/CB =1/2 and the point D is such that DA/BA = 3 and DB/BA = 2. If C = (0, 4), D = (4, 0) and A = (x,y) what is the value of 2x+y?

Apr 29, 2019

### 1+0 Answers

#1
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Let us have four distinct collinear points A, B, C and D on the Cartesian plane.

The point C is such that AB/CB =1/2 and

the point D is such that DA/BA = 3 and

DB/BA = 2.

If C = (0, 4), D = (4, 0) and A = (x,y) what is the value of 2x+y?

$$\text{Let AB=BA} \\ \text{Let DB=BD} \\ \text{Let \vec{C}=\dbinom{0}{4} } \\ \text{Let \vec{D}=\dbinom{4}{0} }$$

$$\begin{array}{|rcll|} \hline \dfrac{AB}{CB} &=& \dfrac{1}{2} \\ \mathbf{ CB } &=& \mathbf{2AB} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \dfrac{DB}{BA} &=& 2\\ \mathbf{ DB } &=& \mathbf{2AB} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline CD &=& CB + BD \\ CD &=& 2AB+2AB \\ \mathbf{ CD } &=& \mathbf{4AB} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \dfrac{DA}{BA} &=& 3 \\ \mathbf{ DA } &=& \mathbf{3AB} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \dfrac{DA}{CD} &=& \dfrac{3AB}{4AB} \\ \mathbf{\lambda_A=\dfrac{DA}{CD} }&=& \mathbf{\dfrac{3}{4}} \\ \hline \end{array}$$

Line:

$$\begin{array}{|rcll|} \hline \vec{x} &=& \vec{D}+\lambda(\vec{C}-\vec{D}) \\ \vec{A} &=& \vec{D}+\lambda_A(\vec{C}-\vec{D}) \quad | \quad \lambda_A = \dfrac{3}{4} \\ \vec{A} &=& \vec{D}+\dfrac{3}{4}(\vec{C}-\vec{D}) \\ &=& \dbinom{4}{0}+\dfrac{3}{4}\left( \dbinom{0}{4}-\dbinom{4}{0} \right) \\\\ &=& \dbinom{4}{0}+\dfrac{3}{4} \dbinom{-4}{4} \\\\ &=& \dbinom{4}{0}+ \dbinom{\dfrac{3}{4}\cdot(-4)}{\dfrac{3}{4}\cdot 4} \\\\ &=& \dbinom{4}{0}+ \dbinom{-3} {3} \\\\ &=& \dbinom{4-3}{3} \\\\ \mathbf{\vec{A} } &=& \mathbf{ \dbinom{1}{3} } \\\\ \vec{A} &=& \dbinom{x}{y} \qquad x = 1,\ y=3 \\ 2x+y &=& 2\cdot 1 + 3 \\ \mathbf{2x+y } &=& \mathbf{ 5 } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \dfrac{DB}{CD} &=& \dfrac{2AB}{4AB} \\ \mathbf{\lambda_B=\dfrac{DB}{CD} }&=& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}$$

Line:

$$\begin{array}{|rcll|} \hline \vec{x} &=& \vec{D}+\lambda(\vec{C}-\vec{D}) \\ \vec{B} &=& \vec{D}+\lambda_B(\vec{C}-\vec{D}) \quad | \quad \lambda_B = \dfrac{1}{2} \\ \vec{B} &=& \vec{D}+\dfrac{1}{2}(\vec{C}-\vec{D}) \\ &=& \dbinom{4}{0}+\dfrac{1}{2}\left( \dbinom{0}{4}-\dbinom{4}{0} \right) \\\\ &=& \dbinom{4}{0}+\dfrac{1}{2} \dbinom{-4}{4} \\\\ &=& \dbinom{4}{0}+ \dbinom{\dfrac{1}{2}\cdot(-4)}{\dfrac{1}{2}\cdot 4} \\\\ &=& \dbinom{4}{0}+ \dbinom{-2} {2} \\\\ &=& \dbinom{4-2}{2} \\\\ \mathbf{\vec{B} } &=& \mathbf{ \dbinom{2}{2} } \\ \hline \end{array}$$

Apr 30, 2019
edited by heureka  Apr 30, 2019
edited by heureka  Apr 30, 2019