Răspuns:
Explicație pas cu pas:
ecuatia unei drepte
y = ax + b
a)
3 = 2a + b
-2 = -a + b
a = b + 2
3 = 2*(b + 2) + b
3 = 2b + 4 + b
3b = 3 - 4 = -1
b = -1/3
a = -1/3 + 2 = -1/3 + 6/3 = 5/3
y = 5x/3 - 1/3
b)
x = (xA + xC)/2 = (2 + 4)/2 = 3
y = (yA + yC)/2 = (3 - 3)/2 = 0
M(3, 0)
c)
AC = √[(2 - 4)² + (3 + 3)²] = √(4 + 36) = √40 = 2√10
[tex]\it a)\ \ (AB):\ \ \dfrac{y-y_A}{y_B-y_A}=\dfrac{x-x_A}{x_B-x_A} \Rightarrow \dfrac{y-3}{-2-3}=\dfrac{x-2}{-1-2} \Rightarrow \dfrac{y-3}{-5}=\dfrac{x-2}{-3} \Rightarrow\\ \\ \\ \Rightarrow y-3=\dfrac{5}{3}x-\dfrac{10}{3}\bigg|_{+3} \Rightarrow y=\dfrac{5}{3}x-\dfrac{1}{3}[/tex]
[tex]\it b)\ \ Fie\ M(x,\ y)\ mijlocul\ lui\ [AC].\\ \\ x=\dfrac{x_A+x_C}{2}=\dfrac{2+4}{2}=3;\ \ \ y=\dfrac{y_A+y_C}{2}=\dfrac{3-3}{2}=0[/tex]
[tex]\it c)\ \ AC^2=(x_C-x_A)^2+(y_C-y_A)^2=(4-2)^2+(-3-3)^2=4+36=40\\ \\ AC=\sqrt{40}=\sqrt{4\cdot10}=2\sqrt{10}[/tex]