Notes: `(2,0), (0,2), (-1,1)` Use calculus to find the area of the triangle with the given vertices.

Monday, 25 September 2017

$\displaystyle{\left({2},{0}\right)},{\left({0},{2}\right)},{\left(-{1},{1}\right)}$ Use calculus to find the area of the triangle with the given vertices.

Given the coordinates (2, 0), (0, 2), and (-1, 1).

Let A=(2, 0), B(0, 2), and C(-1, 1).

Find the equation of line AB using A(2, 0) and B(0, 2).

The slope of line AB is $\displaystyle\frac{{{2}-{0}}}{{{0}-{2}}}=-{1}$

The equation of line AB is $\displaystyle{y}=-{1}{x}+{2}$

Find the equation of line BC using B(0, 2) and C(-1 1).

The slope of line BC is $\displaystyle\frac{{{1}-{2}}}{{-{1}-{0}}}=\frac{{-{1}}}{{-{{1}}}}={1}$

The equation of line BC is $\displaystyle{y}={1}{x}+{2}$

Find the equation of line AC using A(2,...

Given the coordinates (2, 0), (0, 2), and (-1, 1).

Let A=(2, 0), B(0, 2), and C(-1, 1).

Find the equation of line AB using A(2, 0) and B(0, 2).

The slope of line AB is $\displaystyle\frac{{{2}-{0}}}{{{0}-{2}}}=-{1}$

The equation of line AB is $\displaystyle{y}=-{1}{x}+{2}$

Find the equation of line BC using B(0, 2) and C(-1 1).

The slope of line BC is $\displaystyle\frac{{{1}-{2}}}{{-{1}-{0}}}=\frac{{-{1}}}{{-{{1}}}}={1}$

The equation of line BC is $\displaystyle{y}={1}{x}+{2}$

Find the equation of line AC using A(2, 0) and C(-1, 1).

The slope of line AC is $\displaystyle\frac{{{1}-{0}}}{{-{1}-{2}}}=-\frac{{1}}{{3}}$

The equation of line AC is $\displaystyle{y}-{0}={\left(-\frac{{1}}{{3}}\right)}{\left({x}-{2}\right)}\Rightarrow{y}=-\frac{{1}}{{3}}{x}+\frac{{2}}{{3}}$

Set up the intervals for integration.

$\displaystyle{\int_{{-{{1}}}}^{{0}}}{\left({x}+{2}\right)}-{\left(-\frac{{1}}{{3}}{x}+\frac{{2}}{{3}}\right)}{\left.{d}{x}\right.}+{\int_{{0}}^{{2}}}{\left(-{x}+{2}\right)}-{\left(-\frac{{1}}{{3}}{x}+\frac{{2}}{{3}}\right)}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{-{{1}}}}^{{0}}}{\left(\frac{{4}}{{3}}{x}+\frac{{4}}{{3}}\right)}{\left.{d}{x}\right.}+{\int_{{0}}^{{2}}}{\left(-\frac{{2}}{{3}}{x}+\frac{{4}}{{3}}\right)}{\left.{d}{x}\right.}$

$\displaystyle={\left[\frac{{4}}{{3}}\cdot\frac{{{x}}^{{2}}}{{2}}+\frac{{4}}{{3}}{x}\right]}$ from x=-1 to x=0 + $\displaystyle{\left[-\frac{{2}}{{3}}\cdot\frac{{{x}}^{{2}}}{{2}}+\frac{{4}}{{3}}{x}\right]}$ from x=0 to x=2

$\displaystyle={\left[\frac{{2}}{{3}}{{x}}^{{2}}+\frac{{4}}{{3}}{x}\right]}$ from x=-1 to x=0 + $\displaystyle{\left[-\frac{{1}}{{3}}{{x}}^{{2}}+\frac{{4}}{{3}}{x}\right]}$ from x=0 to x=2