Question

qual e a equaçao da reta que passa os pontos (1,3) e (2,4)

Answer 1

$a~(1~;~3)~~~~e~~~~b~(2~;~4)\\\\\\ Calculando~~o~~Coeficiente~~Angular~~m:\\\\ m= \dfrac{y_b-y_a}{x_b-x_a}\to~~ m= \dfrac{4-3}{2-1}\to~~ m= \dfrac{1}{1} \to~~ \boxed{m=1}\\\\\\ (y-y_{\°})=m(x-x_{\°})\\\\\\ Usando~~o~~ponto~~a~(1~;~3):\\\\ (y-3)=1(x-1)\to~~ y-3=x-1\to~~y-x-3+1=0\to\\\\ \boxed{ y-x-2=0} \to~~~~ Equa\c c\~ao~~Geral\\\\\\ \boxed{ y = x+2} \to ~~~~ Equa\c c\~ao~~Reduzida$

Answer 2

Na primeira coordenada
x+y=3
1+y=3
y=3-1
y=2

então podemos definir a seguinte equação em função de x

f(x)= x+2