, 2012

The slope of the line is $m=-3$; thus its equation is $y=-3x+b$ with $y$-intercept $b$. Since the line passes through the point $(2,6)$, then $x=2$ and $y=6$ satisfy the equation of the line. Substituting $x=2$ and $y=6$ into the equation of the line, then $6=-3(2)+b$ and so $b=12$. The equation of the line is $y=-3x+12$ and the line has $y$-intercept 12. To find the $x$-intercept, we let $y=0$ and solve for $x$. Thus, $0=-3x+12$ or $3x=12$, and so the line has $x$-intercept 4. The slope of the line is $m$; thus its equation is $y=mx+b$ with $y$-intercept $b$. Since the line passes through the point $(2,6)$, then $x=2$ and $y=6$ satisfy the equation of the line. Substituting $x=2$ and $y=6$ into the equation of the line, then $6=2m+b$ and so $b=6-2m$. The equation of the line is $y=mx+(6-2m)$ and the line has $y$-intercept $6-2m$. To find the $x$-intercept, we let $y=0$ and solve for $x$. Thus, $0=mx+(6-2m)$ or $mx=2m-6$ or $x=\dfrac{2m-6}{m}$, and so the line has $x$-intercept $2-\dfrac{6}{m}$. (We require $m\neq0$, otherwise the line is horizontal and the $x$-intercept does not exist.) The line through the point $(2,6)$ with slope $m$ has $x$-intercept $2-\dfrac{6}{m}$ and $y$-intercept $6-2m$, as determined in part (b). (We require $m\neq0$, otherwise the line is horizontal and the $x$-intercept, $P$, does not exist.) Since $P$ is the $x$-intercept of this line, $OP$ has length $2-\dfrac{6}{m}$. Since $Q$ is the $y$-intercept of this line, $OQ$ has length $6-2m$. Therefore, the area of $\triangle POQ$ is given by $\dfrac{1}{2}(OP)(OQ)=\dfrac{1}{2}\left(2-\dfrac{6}{m}\right)(6-2m)$. Since the area of $\triangle POQ$ is 25, then $\dfrac{1}{2}\left(2-\dfrac{6}{m}\right)(6-2m)=25$. Solving for $m$, $$\begin{aligned} \frac{1}{2}\left(2-\frac{6}{m}\right)(6-2m)&=25\\ \left(2-\frac{6}{m}\right)(6-2m)&=50\\ (2m-6)(6-2m)&=50m\\ 12m-4m^2-36+12m&=50m\\ 4m^2+26m+36&=0\\ 2m^2+13m+18&=0\\ (2m+9)(m+2)&=0\end{aligned}$$ Therefore, two possible values are $m=-\dfrac{9}{2}$ and $m=-2$. Since $P$ and $Q$ lie on the positive $x$-axis and the positive $y$-axis respectively, we must check that these two values for $m$ give $2-\dfrac{6}{m}>0$ and $6-2m>0$. When $m=-\dfrac{9}{2}$, $2-\dfrac{6}{m}=2+\dfrac{6}{\frac{9}{2}}$ which is greater than 0. When $m=-\dfrac{9}{2}$, $6-2m=6+2(\frac{9}{2})$ which is also greater than 0. When $m=-2$, $2-\dfrac{6}{m}=2+\dfrac{6}{2}$ which is greater than 0. When $m=-2$, $6-2m=6+2(2)$ which is also greater than 0. Therefore, the two values of $m$ for which $P$ and $Q$ lie on the positive $x$-axis and the positive $y$-axis, respectively, and for which $\triangle POQ$ has area 25, are $m=-\dfrac{9}{2}$ and $m=-2$. Note: If we remove the restriction that $P$ and $Q$ both be located on their respective positive axes, then there are two more values of $m$ for which $\triangle POQ$ has area 25. Can you determine these?