Find the area of the triangle under certain preconditions

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.

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

• My mistake was that at the beginning, I didn't consider $(x_1, y_1)$ to be the name of it. I just used $x, y$ – Don Larynx Jan 29 '15 at 21:45
• I can understand; the notation used in the problem made it a little confusing. – user84413 Jan 29 '15 at 22:15

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 $(1/2)\|\left\langle 0, 0, by - ax\right\rangle\| = (1/2)|by - ax|$.

• But why does the magnitude of the cross product correspond to the area of the parallelogram spanned by the two vectors? That's what this question is about as I see it. – Arthur Jan 28 '15 at 23:42
• @Arthur it's not corresponding to a parallelogram, but to the triangle. The vector $\left\langle x,y,0\right\rangle$ is the base of triangle (oriented from the origin to $(x,y,0)$ in $\Bbb R^3$) and $\left\langle b,a,0\right\rangle$ is the side $\ell$ of the triangle (oriented from the origin to $(b,a,0)$ in $\Bbb R^3$). – kobe Jan 28 '15 at 23:56
• We want half a parallelogram, which is why you divide the cross product by $2$. The cross product is often described as having magnitude corresponding to the area of the parallelogram (and thus twice the triangle), but going from there directly to the formula seems to gloss over a few details, and I believe that those are exactly the details the OP cannot figure out. – Arthur Jan 29 '15 at 0:06
• Yes, I'm aware of that. My point was that I didn't need to introduce a parallelogram to get the cross product. I did add some more details in my edit, and if the OP needs more, then I'll add more. – kobe Jan 29 '15 at 0:10
• I'm not exactly sure about why I divide the cross product by half. A picture would help. – Don Larynx Jan 29 '15 at 1:30

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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.

$$\mbox{In the present case:}\qquad \vec{A}=\vec{0}\,,\quad\vec{B}=\pars{b,a,0}\,,\quad \vec{C}=\pars{x,y,0}$$ 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}