Question

Qual a função f(x)= ax+ b, que passa pelos pontos:

A (0,0) e (-1,3)

B (-1,1) e (0,2

C (-4,2) e (2,5)

Answer 1

$\boxed{f(x) = ax + b}$ representa uma reta.

Nós podemos achar uma equação da reta através do determinante.

$\boxed{D = \left[\begin{array}{ccc}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{array}\right]}$

A)

$P(0, 0) \rightarrow (x_1,y_1) \\ L(-1, 3) \rightarrow (x_2, y_2) \\\\ \\ D = \left[\begin{array}{ccc}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{array}\right] \\\\ D = \left[\begin{array}{ccc}x&y&1\\0&0&1\\-1&3&1\end{array}\right] \rightarrow Resolvendo \ pelo \ m\acute{e}todo \ de \ Sarrus \Downarrow \\ D = x*0*1+y*1*(-1)+1*0*3-(-1)*0*1-3*1*x-1*0*y \\ D = 0-y+0+0-3x-0 \\ \boxed{D = -3x-y} \\ Equa\c{c}\tilde{a}o \rightarrow \boxed{-3x-y=0}$

B)

$P(-1, 1) \rightarrow (x_1,y_1) \\ L(0, 2) \rightarrow (x_2,y_2) \\\\ D = \left[\begin{array}{ccc}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{array}\right]} \\\\ D = \left[\begin{array}{ccc}x&y&1\\-1&1&1\\0&2&1\end{array}\right]} \\ D = x*1*1+y*1*0+1*(-1)*2-0*1*1-2*1*x-1*(-1)*y \\ D = x + 0 - 2-0-2x+y \\ D = -x+y-2 \\ Equa\c{c}\tilde{a}o \rightarrow \boxed{-x+y-2=0}$

C)

$P(-4, 2) \rightarrow (x_1,y_1) \\ L(2,5) \rightarrow (x_2,y_2) \\\\ D = \left[\begin{array}{ccc}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{array}\right]} \\\\ D = \left[\begin{array}{ccc}x&y&1\\-4&2&1\\2&5&1\end{array}\right]} \\ D = x*2*1+y*1*2+1*(-4)*5-2*2*1-5*1*x-1*(-4)*y \\ D = 2x + 2y-20-4-5x+4y \\ D = -3x + 6y - 24 \\ Equa\c{c}\tilde{a}o \rightarrow \boxed{-3x+6y-24=0}$