Question

Se consideră paralelogramul ABCD cu A(3,–2), BC: x− y+3 , BD: x-3y-3 :
sa se afle:
a) coordonatele vf paralelogramului
b) aria paralelogramului

Answer 1

Explicație pas cu pas:

A(3, –2)

$BC: x - y + 3 = 0 \iff y_{BC} = x + 3$

$BD: x - 3y - 3 = 0 \iff y_{BD} = \frac{1}{3}x - 1$

$y_{BC} = y_{BD} \iff x + 3 = \frac{1}{3}x - 1 \\ 3x + 9 = x - 3 \implies x_{B} = - 6 \\ y = - 6 + 3 \implies y_{B} = - 3 \\ B(-6;-3)$

$AB: \frac{y - y_{B}}{y_{A} - y_{B}} = \frac{x - x_{B}}{x_{A} - x_{B}} \\ \frac{y - ( - 3)}{ - 2 - ( - 3)} = \frac{x - ( - 6)}{3 - ( - 6)} \\ 9y + 27 = x + 6 \implies y_{AB} = \frac{1}{9}x - \frac{7}{3}$

$AD || BC \implies m_{AD} = m_{BC} = 1$

$AD: y - y_{A} = m_{AD} \cdot (x - x_{A}) \\ y - ( - 2) = 1\cdot (x - 3) \\ y + 2 = x - 3 \implies y_{AD} = x - 5$

$y_{AD} = y_{BD} \implies x - 5 = \frac{1}{3}x - 1 \\ 3x - 15 = x - 3 \implies x_{D} = 6 \\ y = 6 - 5 \implies y_{D} = 1 \\ D(6;1)$

$AB || CD \implies m_{AB} = m_{CD} = \frac{1}{9}$

$CD: y - y_{D} = m_{CD} \cdot (x - x_{D}) \\ y - 1 = \frac{1}{9} \cdot (x - 6) \implies y_{CD} = \frac{1}{9}x + \frac{1}{3}$

$y_{BC} = y_{CD} \implies x + 3 = \frac{1}{9}x + \frac{1}{3} \\ 9x + 27 = x + 3 \implies x_{C} = - 3 \\ y = - 3 + 3 \implies y_{C} = 0 \\ C(-3;0)$

$Aria_{(ABCD)} = 2 \cdot Aria_{(BCD)} = 2 \cdot \frac{1}{2} \cdot \Big|\Delta\Big|$

$\left|\begin{array}{ccc} - 6& - 3&1\\ - 3&0&1\\6&1&1\end{array}\right| = - 24$

$\implies Aria_{(ABCD)} = 24$