At what point does the graph of the parametric equations

x = 3-2t, y = -2 + 5t, intersect the graph of the parametric equations

x = -1+3s,y = 5-9s? The answer is the point of intersection (x,y).

Guest May 6, 2019

#1**+1 **

\(\text{probably easiest to un parameterise them and solve}\\ x = 3-2t \Rightarrow t = \dfrac{3-x}{2}\\ y = -2+5\cdot \dfrac{3-x}{2} = -2+\dfrac{15}{2}-\dfrac 5 2 x = -\dfrac 5 2 x + \dfrac{11}{2}\)

\(x = -1+3s \Rightarrow s = \dfrac{x+1}{3}\\ y = 5 - 9\cdot \dfrac{x+1}{3} = 5-3x-3= -3x+2\)

\(-\dfrac 5 2 x + \dfrac{11}{2}=-3x+2\\ \dfrac 1 2 x = -\dfrac 7 2\\ x = -7\\ y = 23\\ (x,y)=(-7,23)\)

.Rom May 6, 2019

#2**+2 **

**At what point does the graph of the parametric equations**

**x = 3-2t, y = -2 + 5t, intersect the graph of the parametric equations**

**x = -1+3s,y = 5-9s? The answer is the point of intersection (x,y).**

\(\begin{array}{|lrcll|} \hline 1) & 3-2t &=& -1+3s \\ & 3s+2t &=& 4 \quad | \quad \cdot 3 \\ &\mathbf{ 9s+6t} &=& \mathbf{12} \qquad (1) \\\\ 2) & -2+5t &=& 5-9s \\ & 9s+5t &=& 7 \\ &\mathbf{ 9s+5t} &=& \mathbf{7} \qquad (2) \\ \hline (1)-(2): & 9s+6t -9s-5t &=& 12-7 \\ &\mathbf{t} &=& \mathbf{5} \\ \hline \text{The point of intersection } (x,y)\\ & x &=& 3-2t \\ & x &=& 3-2\cdot 5 \\ & \mathbf{x} &=& \mathbf{-7} \\\\ & y &=& -2+5t \\ & y &=& -2+5\cdot 5 \\ & \mathbf{y} &=& \mathbf{23} \\ \hline & \mathbf{(x,y)} &=& \mathbf{(-7,\ 23)} \\ \hline \end{array}\)

heureka May 6, 2019