Find the area of the triangle under certain preconditions

Question

With vertices $(0, 0)$, $(b, a)$, $(x, y)$, prove the area of this triangle is $\frac{|by - ax|}{2}$.

We know area of a triangle = $\frac{rh}{2}$. ($r$ is the base.) Well, we have $r =$ the distance from the origin to $(x, y)$ and the height is the line perpendicular to $y = mx$, which is $y = -\frac{x}{m} + b'$, intersecting the points $(x_0, y_0)$ on the line $y = mx$ and $(b, a)$.

So then, the altitude from my calculations was $ h =\sqrt{1+\frac{1}{m^2}}|x_0 - b|$ The base is just the distance to the origin from $(x, y)$, which is $d = \sqrt{x^2+y^2}$. But I don't see how this equals what I'm trying to prove.

Answer 1

If we denote the point $(x,y)$ by $(x_1,y_1)$ to avoid confusion, we can take the base of the triangle to be $\;\;\;B=\sqrt{x_1^2+y_1^2}$.

The line through $(0,0)$ and $(x_1,y_1)$ has equation $y=\frac{y_1}{x_1}x$ (assuming $x_1\ne0$), so the perpendicular line through $(b,a)$ has equation $y-a=-\frac{x_1}{y_1}(x-b)$ or $y=-\frac{x_1}{y_1}(x-b)+a$ (assuming $y_1\ne0$).

The two lines intersect where $\frac{y_1}{x_1}x=-\frac{x_1}{y_1}(x-b)+a$, so solving for $x$ gives

$\left(\frac{y_1}{x_1}+\frac{x_1}{y_1}\right)x=\frac{x_1 b}{y_1}+a$, so $x=\frac{x_1y_1}{x_1^2+y_1^2}\left(\frac{x_1 b}{y_1}+a\right)\implies x=\frac {x_1(bx_1+ay_1)}{x_1^2+y_1^2}\text{ and }y=\frac{y_1(bx_1+ay_1)}{x_1^2+y_1^2}.$

Then the height of the triangle is given by the distance from $(b,a)$ to this point, so

$\displaystyle h=\sqrt{\left(b-\frac {x_1(bx_1+ay_1)}{x_1^2+y_1^2}\right)^2+\left(a-\frac{y_1(bx_1+ay_1)}{x_1^2+y_1^2}\right)^2}$

$\displaystyle=\sqrt{\frac{y_1^{2}(b y_1-a x_1)^{2}+x_1^{2}(ax_1-by_1)^2}{(x_1^{2}+y_1^{2})^2}}=\sqrt{\frac{(by_1-ax_1)^2}{x_1^2+y_1^2}}=\frac{\lvert by_1-ax_1\rvert}{\sqrt{x_1^2+y_1^2}}$.

Then $\displaystyle A=\frac{1}{2}Bh=\frac{|by_1-ax_1|}{2}$.

Answer 2

Perhaps you'll find it easier to compute the area by expressing it as half the magnitude of the vector product $\left\langle b,a,0\right\rangle \times \left\langle x,y,0\right\rangle$. Indeed, if $\theta$ is the interior angle between the base and the side $\ell$ from the origin to $(x,y)$, then $h = \ell\sin \theta$. So the area is $(1/2)r\ell\sin \theta$. This is the same as $(1/2)\|\left\langle b,a,0\right\rangle \times \left\langle x,y,0\right\rangle\|$. The cross product equals $\left\langle 0, 0, by - ax \right\rangle$. Therefore, the area is $(1/2)\|\left\langle 0, 0, by - ax\right\rangle\| = (1/2)|by - ax|$.

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ If $\ds{\vec{A},\ \vec{B}\ \mbox{and}\ \vec{C}}$ are the triangles vertices, the area $\ds{\,{\cal A}}$ is given by: \begin{align} \,{\cal A}=\half\,\verts{\vec{A}\times\vec{B}} +\half\,\verts{\vec{B}\times\vec{C}} + \half\,\verts{\vec{C}\times\vec{A}}\tag{1} \end{align} The proof is simple as it involves the Stokes Theorem.

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$$ \mbox{In the present case:}\qquad \vec{A}=\vec{0}\,,\quad\vec{B}=\pars{b,a,0}\,,\quad \vec{C}=\pars{x,y,0} $$ Then, with formula $\pars{1}$: \begin{align} \color{#66f}{\large\,{\cal A}}&=\half\,\verts{\vec{B}\times\vec{C}} =\half\verts{\pars{b,a,0}\times\pars{x,y,0}} =\half\verts{\vphantom{\Huge A^{A^{A^{A}}}}\verts{% \begin{array}{ccc}{\bf i} & {\bf j} & {\bf k} \\ b & a & 0 \\ x & y & 0 \end{array}}}=\half\,\verts{\pars{by - ax}{\bf k}} \\[5mm]&=\color{#66f}{\large\frac{\verts{by - ax}}{2}} \end{align}