, 2018

We determine the $x$-intercept by letting $y=0$ in the equation $y=2x-6$ and solving for $x$. Thus, $0=2x-6$ and so $2x=6$ or $x=3$. The $x$-intercept of the line with equation $y=2x-6$ is 3. We determine the $y$-intercept by letting $x=0$ in the equation $y=2x-6$ and solving for $y$. Thus, $y=2(0)-6$ and so $y=-6$. The $y$-intercept of the line with equation $y=2x-6$ is $-6$. Letting $y=0$, we get $0=kx-6$ or $kx=6$ and so $x=\dfrac{6}{k}$, where $k\neq0$. The line with equation $y=kx-6$ has $x$-intercept $\dfrac{6}{k}$ ($k\neq0$). From part (b), the line with equation $y=kx-6$ has $x$-intercept $\dfrac{6}{k}$. Since $k>0$, then $\dfrac{6}{k}>0$ and so the line intersects the positive $x$-axis. The $y$-intercept of the line with equation $y=kx-6$ is $-6$. [[IMAGE0]] The triangle formed by the line with equation $y=kx-6$ ($k>0$), the positive $x$-axis, and the negative $y$-axis, has area $\dfrac12\left(\dfrac6k\right)(6)=\dfrac{36}{2k}=\dfrac{18}{k}$ (the $y$-intercept is $-6$, and so the triangle has height $6$). Since the area of this triangle is 6, then $\dfrac{18}{k}=6$ or $18=6k$ and so $k=3$. The $x$-intercept of the line with equation $y=2mx-m^2$ is determined by letting $y=0$ and solving for $x$. Thus, $0=2mx-m^2$ or $0=m(2x-m)$ and since $m>0$, then $2x=m$ or $x=\dfrac{m}{2}$. The $x$-intercept of this line is $\dfrac{m}{2}$ ($m>0$). The $y$-intercept of the line with equation $y=2mx-m^2$ is $2m(0)-m^2=-m^2$. Similarly, the $x$-intercept of the line with equation $y=mx-m^2$ is given by $0=mx-m^2$ or $0=m(x-m)$ and since $m>0$, then $x=m$. The $x$-intercept of this line is $m$ ($m>0$). The $y$-intercept of the line with equation $y=mx-m^2$ is $m(0)-m^2=-m^2$. Thus, both lines have the same $y$-intercept. [[IMAGE1]] To determine the area of the triangle formed by the positive $x$-axis, the line with equation $y=mx-m^2$, and the line with equation $y=2mx-m^2$ ($m>0$), we may let the length of the base be the distance between the $x$-intercepts or $m-\dfrac{m}{2}=\dfrac{m}{2}$. Then the height of this triangle is the perpendicular distance from the $x$-axis to the $y$-intercept, or $m^2$ (the $y$-intercept is $-m^2$, and so the triangle has height $m^2$, a positive number). Therefore, the triangle has area $\dfrac12\left(\dfrac{m}{2}\right)(m^2)=\dfrac{m^3}{4}$. The area of this triangle is $\dfrac{54}{125}$ and so $\dfrac{m^3}{4}=\dfrac{54}{125}$ or $m^3=\dfrac{216}{125}$ and so $m=\sqrt[3]{\dfrac{216}{125}}=\dfrac{6}{5}$. (Note that $\left(\dfrac{6}{5}\right)^3=\dfrac{216}{125}$.) The only value of $m$ for which the triangle has area $\dfrac{54}{125}$ is $m=\dfrac65$.