Question

2. Se dau punctele A(-2;-2), B(1;0), C(3;5), D(-1;4) a) Reprezentați punctele în repreul cartezian xoy b) Scrieți ecuațiile laturilor patrulaterului ABCD c) Scrieți ecuațiile diagonalelor patrulaterului ABCD d) Comparați aria triunghiului ABC cu aria triunghiului BDC​

Answer 1

Explicație pas cu pas:

a) A(-2;-2), B(1;0), C(3;5), D(-1;4)

b)

$\frac{y - y_{B}}{y_{A} - y_{B}} = \frac{x - x_{B}}{x_{A} - x_{B}} = \frac{y - 0}{-2 - 0} = \frac{x - 1}{-2 - 1} \\ \frac{y}{- 2} = \frac{x - 1}{ - 3} < = > 3y = 2x - 2 \\ = > 2x - 3y - 2 = 0$

$AB: 2x - 3y - 2 = 0$

$\frac{y - y_{C}}{y_{B} - y_{C}} = \frac{x - x_{C}}{x_{B} - x_{C}} = \frac{y - 5}{0 - 5} = \frac{x - 3}{1 - 3} \\ \frac{y - 5}{ - 5} = \frac{x - 3}{ - 2} < = > 2(y - 5) = 5(x - 3) \\ 2y - 10 = 5x - 15 = > 5x - 2y - 5 = 0$

$BC: 5x - 2y - 5 = 0$

$\frac{y - y_{D}}{y_{C} - y_{D}} = \frac{x - x_{D}}{x_{C} - x_{D}} = \frac{y - 4}{5 - 4} = \frac{x - (-1)}{3 - (-1)} \\ \frac{y - 4}{1} = \frac{x + 1}{4} < = > 4(y - 4) = x + 1 \\ 4y - 16 = x + 1 = > x - 4y + 17 = 0$

$CD: x - 4y + 17 = 0$

$\frac{y - y_{D}}{y_{A} - y_{D}} = \frac{x - x_{D}}{x_{A} - x_{D}} = \frac{y - 4}{-2 - 4} = \frac{x - (-1)}{-2 - (-1)} \\ \frac{y - 4}{ - 6} = \frac{x + 1}{- 1} < = > y - 4 = 6(x + 1) \\ y - 4 = 6x + 6 = > 6x - y + 10 = 0$

$AD: 6x - y + 10 = 0$

c)

$\frac{y - y_{C}}{y_{A} - y_{C}} = \frac{x - x_{C}}{x_{A} - x_{C}} = \frac{y - 5}{-2 - 5} = \frac{x - 3}{-2 - 3} \\ \frac{y - 5}{ - 7} = \frac{x - 3}{ - 5} < = > 5(y - 5) = 7(x - 3) \\ 5y - 25 = 7x - 21 = > 7x - 5y + 4 = 0$

$AC: 7x - 5y + 4 = 0$

$\frac{y - y_{D}}{y_{B} - y_{D}} = \frac{x - x_{D}}{x_{B} - x_{D}} = \frac{y - 4}{0 - 4} = \frac{x - (- 1)}{1 - ( - 1)} \\ \frac{y - 4}{ - 4} = \frac{x + 1}{2} < = > y - 4 = - 2(x + 1) \\ y - 4 = - 2x - 2 = > 2x + y - 2 = 0$

$BD: 2x + y - 2 = 0$

d) A(-2;-2), B(1;0), C(3;5), D(-1;4)

$\left|\begin{array}{ccc}-2&-2&1\\1&0&1\\3&5&1\end{array}\right| = 11$

$Aria_{(ABC)} = \frac{1}{2} \times |11| = \frac{11}{2}$

$\left|\begin{array}{ccc}1&0&1\\-1&4&1\\3&5&1\end{array}\right| = - 18$

$Aria_{(BDC)} = \frac{1}{2} \times | - 18| = 9$

→

$Aria_{(ABC)} < Aria_{(BDC)}$