Sos câu 5 với mn sos

Đáp án: $=\dfrac{49}3$

Giải thích các bước giải:  

Ta có:

$\dfrac34x-2=\dfrac45x-3\to x=20\to y=\dfrac34\cdot 20-2=13\to A(20, 13)$

$\dfrac45x-3=\dfrac12x-4\to x=-\dfrac{10}3\to y=\dfrac45\cdot (-\dfrac{10}3)-3=-\dfrac{17}{3}\to B(-\dfrac{10}3, -\dfrac{17}{3})$

$\dfrac12x-4=\dfrac34x-2\to x=-8\to y=\dfrac12\cdot (-8)-4=-8\to C(-8, -8)$

Ta có: $AC$ là đường thẳng $(d_1)$

$\to (AC): y=\dfrac34x-2$

Kẻ $BH\perp AC$

$\to BH: y=-\dfrac43(x+\dfrac{10}3)-\dfrac{17}3\to y=\dfrac{-12x-91}{9}$

Giải $\dfrac{-12x-91}{9}=\dfrac34x-2\to x=-\dfrac{292}{75}$

$\to y= -\dfrac{123}{25}$

$\to H(-\dfrac{292}{75}, -\dfrac{123}{25})$

$\to BH=\sqrt{(-\dfrac{292}{75}-(-\dfrac{10}3))^2+(-\dfrac{123}{25}-(-\dfrac{17}{3}))^2}=\dfrac{14}{15}$

      $AC=\sqrt{(20+8)^2+(13+8)^2}=35$

$\to S_{ABC}=\dfrac12\cdot \dfrac{14}{15}\cdot 35=\dfrac{49}3$