, 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 $\text{k gt;0}$, then $\text{6 k gt;0}$ and so the line intersects the positive $x$-axis.

The $y$-intercept of the line with equation $y=kx-6$ is $-6$.

The triangle formed by the line with equation $y=kx-6$ ($\text{k gt;0}$), 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 $\text{m gt;0}$, then $2x=m$ or $x=\dfrac{m}{2}$.

The $x$-intercept of this line is $\dfrac{m}{2}$ ($\text{m gt;0}$).

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 $\text{m gt;0}$, then $x=m$.

The $x$-intercept of this line is $m$ ($\text{m gt;0}$).

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.

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 $y=2mx-m^2$ ($\text{m gt;0}$), 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 $m=\dfrac65$.