Question

encontre a equação reduzida da reta que passa pelos pontos: A(3,-2) e (-3,-1) B(2,-3)e (-4,3) C(-1,4)e (-4,3) D(5,2)e (-2,-3) F(5,0)e (-1,-4)​

Answer 1

Resposta:

Sem calculo, porém confiança . .

$y=mx+n\\(y-y_{0})=m*(x-x_{0})$

$a)\boxed{y=-\frac{x}{6}-\frac{3}{2}}\\\\b)\boxed{y=-x-1}\\\\c)\boxed{y=\frac{x+13}{3}}\\\\d)\boxed{y=\frac{5x-11}{7}}\\\\e)\boxed{y=\frac{2x-10}{3}}$

Por determinante, letra a;

$\left[\begin{array}{ccccc}x&y&1&x&y&3&-2&1&3&-2&-3&-1&1&-3&-1\end{array}\right]=0\\\\-2x-3y-3-3y+x-6=0\\-x-6y-9=0\\-x-9=6y\\6y=-x-9\\y=\frac{-x-9}{6}\\a)\boxed{y=-\frac{x}{6}-\frac{3}{2}}$

Por determinante, letra b;

$\left[\begin{array}{ccccc}x&y&1&x&y&2&-3&1&2&-3&-4&3&1&-4&3\end{array}\right]=0\\\\-3x-4y+6-2y-3x-12=0\\-6x-6y-6=0\\\frac{-6x-6y-6=0}{6}\\-x-y-1=0\\-x-1=y\\\boxed{y=-x-1}$

letra c;

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\m=\frac{3-4}{-4+1}\\m=\frac{-1}{-3}\\m=\frac{1}{3}\\(y-4)=\frac{1}{3}*(x+1)\\y-4=\frac{x+1}{3}\\3y-12=x+1\\\boxed{y=\frac{x+13}{3}}$

letra d;

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\m=\frac{-3-2}{-2-5}\\m=\frac{5}{7}\\\\(y+3)=\frac{5x}{7}+\frac{10}{7}\\7y+21=5x+10\\7y=5x+10-21\\\boxed{y=\frac{5x-11}{7}}$

letra e;

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\m=\frac{-4-0}{-1-5}\\m=\frac{-4}{-6}\\m=\frac{2}{3}\\\\(y-0)=\frac{2x-10}{3}\\\boxed{y=\frac{2x-10}{3}}$