Answer:
Step-by-step explanation:
Given that:
The vector is the straight line, so to form the algebraic description of this vector we need to find out the slope of it:
Slope: [tex]\frac{y2-y1}{x2-x1}[/tex] = [tex]\frac{2-4}{5-2}[/tex] = [tex]\frac{-2}{3}[/tex]
We have the standard form of the linear is:
y = mx +b
In this situation, y = [tex]\frac{-2}{3}[/tex]x + b (1)
Because the line go through the point: (2, 4) so we substitute them into the equation (1): 4 = [tex]\frac{-2}{3}[/tex] (2) + b <=> b = [tex]\frac{16}{3}[/tex]
So the algebraic description of this vector is: y = [tex]\frac{-2}{3}[/tex]x + [tex]\frac{16}{3}[/tex]