Exercise 8: We give the two lines (D): $2 x - y - 40$ and $(D^{'}) : 2 x + 4 y + 1 = 01$ )
Give a normal vector and a direction vector of each of the two lines (D) and (D'). 2) Show that
(D) and (D') are perpendicular to a point I that we will determine its coordinates. 3 $) (D)$ intersects the $x^{'} x$ axis at point $J$ , and (D') intersects the $y^{'} y$ axis at point K. Determine the coordinates of $I$ and $K .4$ ) Calculate the area of the triangle IJK.

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4) Area of triangle IJK: $\frac{1}{2} | (20) (- \frac{63}{5} - - \frac{1}{4}) - (0) (- \frac{63}{5} - 0) | = 120$

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Step 1 ：1) Normal vector of (D): $(\begin{matrix} 2 \\ - 1 \end{matrix})$ , direction vector of (D): $(\begin{matrix} 1 \\ 2 \end{matrix})$ ; normal vector of (D'): $(\begin{matrix} 2 \\ 4 \end{matrix})$ , direction vector of (D'): $(\begin{matrix} - 4 \\ 2 \end{matrix})$

Step 2 ：2) Equating dot products of direction vectors to 0: $(1) (- 4) + (2) (2) = 0$ ; Finding point I: intercept form of (D) and (D'): $y = 2 x - 40$ and $y = - \frac{1}{2} x - \frac{1}{4}$ ; solve for I: $x_{I} = \frac{39}{5}$ , $y_{I} = - \frac{63}{5}$

Step 3 ：3) Coordinates of J: $(20, 0)$ ; Coordinates of K: $(0, - \frac{1}{4})$

Step 4 ：4) Area of triangle IJK: $\frac{1}{2} | (20) (- \frac{63}{5} - - \frac{1}{4}) - (0) (- \frac{63}{5} - 0) | = 120$

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