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# help with coordinates

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A triangle has coordinates (0,0), (1,2), and (3,1).  Find the perimeter and the area of the triangle.

Jul 1, 2020

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A triangle has coordinates A(0,0), C(1,2), and B(3,1).  Find the perimeter and the area of the triangle.

Hello Guest!

The triangle ABC has the sides a, b, c.
a, b and c are sections of linear functions in the rectangular coordinate system.

Then:

$$A=\int_0^1f_b(x)dx+\int_1^3 f_a(x)dx-\int_0^3 f_c(x)dx$$

$$f_b(x)=2x\\ f_c(x)= \frac{1}{3}x$$

$$f_a(x)=\frac{y_B-y_C}{x_B-x_C}(x-x_C)+y_C\\ f_a(x)=\frac{-1}{2}(x-1)+2=- \frac{1}{2}x+\frac{5}{2}\\ \color{blue} f_a(x)=-\frac{1}{2}x+\frac{5}{2}\\$$

$${\color{blue}\int_0^1f_b(x)dx}=\int_0^1(2x)dx=| x^2|_0^1=1-0\color{blue}=1$$

$${\color{blue}\int_1^3 f_a(x)dx}=\int_1^3(-\frac{1}{2}x+\frac{5}{2})dx=|-\frac{1}{4}x^2+\frac{5}{2}x|_1^3\\ =(-\frac{9}{4}+\frac{15}{2})-(-\frac{1}{4}+\frac{5}{2})\color{blue}=3$$

$${\color{blue}\int_0^3 f_c(x)dx}=\int_0^3( \frac{1}{3}x)dx=|\frac{1}{6}x^2 |_0^3=\frac{9}{6}-0\color{blue}=\frac{3}{2}$$

$$A=\int_0^1f_b(x)dx+\int_1^3 f_a(x)dx-\int_0^3 f_c(x)dx=1+3-\frac{3}{2}$$

$$\large A=\frac{5}{2}$$

$$P=a+b+c$$

$$a=\sqrt{(2-1)^2+(3-1)^2}=\sqrt{1+4}=\sqrt{5}\\ b=\sqrt{1^2+2^2}=\sqrt{5}\\ c=\sqrt{1^2+3^2}=\sqrt{10}=\sqrt{2}\cdot\sqrt{5}$$

$$\large {\color{blue} P=a+b+c}=\sqrt{2}\cdot\sqrt{5}+2\sqrt{5}=\color{blue}\sqrt{5}\cdot(\sqrt{2}+2)$$

$$\large P=7.6344$$

!

Jul 1, 2020
edited by asinus  Jul 2, 2020