Question

Сторони трикутника лежать на прямих: .....
Обчислити його площу.

Стороны треугольника лежат на прямых: .....
Вычислить его площадь.

Answer 1

Ответ:

$\displaystyle \frac{12}{7}$

Объяснение:

Координати точки А

$\displaystyle \begin{cases}4x + 5y - 2 = 0\\2x - y - 8 = 0 \end{cases}\\\\\begin{cases}4x + 5y = 2 \\2x - y = 8 \ \ \ |\cdot 5 \end{cases}\\\\\begin{cases}4x + 5y = 2 \\10x - 5y = 40 \end{cases}$

+____________________

$\displaystyle 14x=42\ \ \ |:14\\\\x=3\\\\\\2x - y = 8\\\\2\cdot 3 - y = 8\\\\6 - y = 8\\\\y=6-8\\\\y=-2\\\\A(3;-2)$

Координати точки B

$\displaystyle \begin{cases}x - 4y - 5 = 0\\4x + 5y - 2 = 0 \end{cases}\\\\\begin{cases}x - 4y = 5\ \ \ |\cdot (-4)\\4x + 5y =2 \end{cases}\\\\\begin{cases}-4x+16y = -20\\4x + 5y =2 \end{cases}$

+____________________

$\displaystyle 21y=-18\ \ \ |:21\\\\y=-\frac{6}{7}\\\\\\x - 4y = 5\\\\x - 4\cdot(-\frac{6}{7}) = 5\\\\x+\frac{24}{7} = 5\\\\x = 5-\frac{24}{7}\\\\x=\frac{11}{7}\\\\B(\frac{11}{7};-\frac{6}{7})$

Координати точки C

$\displaystyle \begin{cases}x - 4y - 5 = 0\\2x - y - 8 = 0 \end{cases}\\\\\begin{cases}x - 4y = 5\\2x - y = 8\ \ \ |\cdot (-4) \end{cases}\\\\\begin{cases}x - 4y = 5\\-8x +4y =-32 \end{cases}$

+____________________

$\displaystyle -7x=-27\ \ \ |:(-7)\\\\x=\frac{27}{7}\\\\\\2x - y = 8\\\\2\cdot\frac{27}{7} - y = 8\\\\\frac{54}{7} - y = 8\\\\y =\frac{54}{7} - 8\\\\y=-\frac{2}{7}\\\\C(\frac{27}{7};-\frac{2}{7})$

Довжина BC

$\displaystyle BC=\sqrt{(\frac{27}{7}-\frac{11}{7})^2+(-\frac{2}{7}+\frac{6}{7})^2}\\\\BC=\sqrt{(\frac{16}{7})^2+(\frac{4}{7})^2}\\\\BC=\sqrt{\frac{256}{49}+\frac{16}{49}}\\\\BC=\sqrt{\frac{272}{49}}\\\\BC=\frac{4\sqrt{17}}{7}$

Висота трикутника

$\displaystyle x - 4y - 5 = 0\\\\A(3;-2)\\\\\\h = \frac{|Ax_0 +By_0 + C|}{\sqrt{A^2 +B^2}}\\\\h = \frac{|1\cdot3 +\left(-4\right) \cdot\left( -2\right) + \left(-5 \right) |}{\sqrt{1^2 +\left( -4\right) ^2}}\\\\h = \frac{| 3+ 8 -5 |}{\sqrt{1 + 16 }}\\\\h = \frac{| 6 |}{\sqrt{17}}\\\\h= \frac{6}{\sqrt{17}}\\\\h= \frac{6\sqrt{ 17}}{ 17}$

Площа трикутника

$\displaystyle S=\frac{BCh}{2}\\\\S=\frac{\frac{4\sqrt{17}}{7}\cdot\frac{6\sqrt{ 17}}{ 17} }{2}\\\\S=\frac{24 }{14}\\\\S=\frac{12 }{7}$