Question

ME AJUDEM

Seja A a intersecção das retas r, de equação 4x –y – 2 = 0 , e s, de equação 2x – y = 0. Se B e C são as intersecções respectivas dessas retas com o eixo das abscissas, a área do triângulo ABC é?

Answer 1

Explicação passo-a-passo:

• Ponto A

$\sf r:~4x-y-2=0$

$\sf y=4x-2$

$\sf s:~2x-y=0$

$\sf y=2x$

Igualando:

$\sf 4x-2=2x$

$\sf 4x-2x=2$

$\sf 2x=2$

$\sf x=\dfrac{2}{2}$

$\sf x=1$

Assim:

$\sf y=2x$

$\sf y=2\cdot1$

$\sf y=2$

Logo, $\sf A(1,2)$

• Ponto B

$\sf 4x-y-2=0$

$\sf 4x-0-2=0$

$\sf 4x=2$

$\sf x=\dfrac{2}{4}$

$\sf x=\dfrac{1}{2}$

Assim, $\sf B\Big(\dfrac{1}{2},0\Big)$

• Ponto C

$\sf 2x-y=0$

$\sf 2x-0=0$

$\sf 2x=0$

$\sf x=\dfrac{0}{2}$

$\sf x=0$

Desse modo, $\sf C(0,0)$

Temos que:

$\sf D=\Big(\begin{array}{ccc} \sf x_A & \sf y_A & \sf 1 \\ \sf x_B & \sf y_B & \sf 1 \\ \sf x_C & \sf y_C & \sf 1 \end{array}\Big)$

$\sf D=\Big(\begin{array}{ccc} \sf 1 & \sf 2 & \sf 1 \\ \\ \sf \frac{1}{2} & \sf 0 & \sf 1 \\ \\ \sf 0 & \sf 0 & \sf 1 \end{array}\Big)$

$\sf det~(D)=1\cdot0\cdot1+2\cdot1\cdot0+1\cdot\dfrac{1}{2}\cdot0-0\cdot0\cdot1-0\cdot1\cdot1-1\cdot\dfrac{1}{2}\cdot2$

$\sf det~(D)=0+0+0-0-0-1$

$\sf det~(D)=-1$

A área do triângulo ABC é:

$\sf A=\dfrac{|~det~(D)~|}{2}$

$\sf A=\dfrac{|~-1~|}{2}$

$\sf A=\dfrac{1}{2}$

$\sf \red{A=0,5~u.a.}$