Blog: A line passes through the point `(10,-2)` and forms with the axes a triangle of area of 9 sq units. Find the equation of the line.

Hello!

Denote the slope of this line as $\displaystyle{m}.$ The vertical line (which has an undefined slope) doesn't suit us, so we'll not miss a solution. Horizontal line with $\displaystyle{m}={0}$ doesn't suit also, so we can divide by $\displaystyle{m}.$

The equation of such a line is $\displaystyle{y}={m}\cdot{\left({x}-{10}\right)}-{2}.$ The triangle formed with this line and the axes is a right one (because the axes are perpendicular to each other). So its area is $\displaystyle\frac{{1}}{{2}}\cdot{\left|{O}{X}\right|}\cdot{\left|{O}{Y}\right|},$ where $\displaystyle{O}$ is the origin, $\displaystyle{X}$ is the x-intercept of the line and $\displaystyle{Y}$ is the y-intercept.

The y-intercept is $\displaystyle{y}{\left({0}\right)}=-{10}{m}-{2}.$ The x-intercept is the $\displaystyle{x}$ for which $\displaystyle{y}={m}\cdot{\left({x}-{10}\right)}-{2}={0},$ so it is $\displaystyle\frac{{2}}{{m}}+{10}.$

Thus our equation for m becomes

$\displaystyle\frac{{1}}{{2}}{\left|{\left(-{10}{m}-{2}\right)}\cdot{\left(\frac{{2}}{{m}}+{10}\right)}\right|}={9}.$

It is the same as $\displaystyle{\left|{\left({5}{m}+{1}\right)}\cdot{\left(\frac{{1}}{{m}}+{5}\right)}\right|}=\frac{{9}}{{2}},$ or $\displaystyle{\left|\frac{{1}}{{m}}{{\left({5}{m}+{1}\right)}}^{{2}}\right|}=\frac{{9}}{{2}}.$

If we suppose $\displaystyle{m}$ is positive, then it becomes

$\displaystyle{{\left({5}{m}+{1}\right)}}^{{2}}=\frac{{9}}{{2}}{m},$ or $\displaystyle{25}{{m}}^{{2}}+{10}{m}+{1}=\frac{{9}}{{2}}{m},$ or $\displaystyle{25}{{m}}^{{2}}+\frac{{11}}{{2}}{m}+{1}={0},$ or $\displaystyle{50}{{m}}^{{2}}+{11}{m}+{2}={0}.$ This equation has no solutions.

Well, what about negative m's? The equation becomes $\displaystyle{{\left({5}{m}+{1}\right)}}^{{2}}=-\frac{{9}}{{2}}{m},$ or $\displaystyle{25}{{m}}^{{2}}+{10}{m}+{1}=-\frac{{9}}{{2}}{m},$ or $\displaystyle{25}{{m}}^{{2}}+\frac{{29}}{{2}}{m}+{1}={0},$ or $\displaystyle{50}{{m}}^{{2}}+{29}{m}+{2}={0}.$

The discriminant is $\displaystyle{D}={{29}}^{{2}}-{4}\cdot{50}\cdot{2}={{29}}^{{2}}-{{20}}^{{2}}={9}\cdot{49},$ so $\displaystyle\sqrt{{{D}}}={3}\cdot{7}={21}.$ The solutions are $\displaystyle\frac{{-{29}\pm{21}}}{{100}},$ $\displaystyle{m}_{{1}}=-\frac{{50}}{{100}}=-\frac{{1}}{{2}},$ $\displaystyle{m}_{{2}}=-\frac{{8}}{{100}}=-\frac{{2}}{{25}}.$ And both are negative as supposed.

Uff. There are two possible equations, $\displaystyle{y}=-\frac{{1}}{{2}}{\left({x}-{10}\right)}-{2}=-\frac{{1}}{{2}}{x}+{3}$ and $\displaystyle{y}=-\frac{{2}}{{25}}{\left({x}-{10}\right)}-{2}=-\frac{{2}}{{25}}{x}-\frac{{6}}{{5}}.$