6. A line has equation $y=k x$, where $k \neq 0$ and $k \neq-1$. The line is reflected in the line with equation $x+y=1$. Determine the slope and the $y$-intercept of the resulting line, in terms of $k$.
6. Every line is determined by any two points on the line.
We use $\mathcal{L}$ to denote the line with equation $x+y=1$.
We use $\ell$ to denote the line with equation $y=k x$ and $\ell^{\prime}$ to denote the line obtained after reflecting $\ell$ in $\mathcal{L}$.
We determine the slope and $y$-intercept of $\ell^{\prime}$ by finding two points on $\ell^{\prime}$ and then using these points to determine the slope, equation, and $y$-intercept.
The first point that we choose is the point, $P$, of intersection between the original line, $\ell$, $y=k x$, and the line of reflection, $\mathcal{L}$.
We note that the line of reflection can be re-written as $y=-x+1$, which has slope -1 .
Since $k \neq-1$, then $\ell$ is not parallel to $\mathcal{L}$ and so there is indeed a point of intersection.
Since $P$ is on $\ell$, then its reflection will be on the resulting line.
Since $P$ is on $\mathcal{L}$, then its reflection will be itself.
In other words, $P$ is also on $\ell^{\prime}$.
To find the $x$-coordinate of $P$, we substitute $y=k x$ into $x+y=1$ to obtain $x+k x=1$ or $x(k+1)=1$ and so $x=\frac{1}{k+1}$.
Since $P$ is on $\ell$, which has equation $y=k x$, then $y=k \cdot \frac{1}{k+1}=\frac{k}{k+1}$.
Thus, the coordinates of $P$ are $\left(\frac{1}{k+1}, \frac{k}{k+1}\right)$.

Consider next the point $O(0,0)$ which is on $\ell$.
Let $Q$ be the reflection of $O$ in $\mathcal{L}$.
Since $Q$ is the reflection of $O$ in $\mathcal{L}$, then $O Q$ is perpendicular to $\mathcal{L}$.
Since $\mathcal{L}$ has slope -1 , then any line perpendicular to $\mathcal{L}$ has slope equal to $-\frac{1}{-1}=1$.
Therefore, $O Q$ has slope 1 .
Since $O$ is the origin $(0,0)$, then the line through $O$ and $Q$ has equation $y=x$.
Thus, $Q$ has coordinates $(q, q)$ for some real number $q$.
In addition to $O Q$ being perpendicular to $\mathcal{L}$, we also need the midpoint of $O Q$ to lie on $\mathcal{L}$.
The midpoint of $O(0,0)$ and $Q(q, q)$ has coordinates $\left(\frac{1}{2}(0+q), \frac{1}{2}(0+q)\right)=\left(\frac{1}{2} q, \frac{1}{2} q\right)$.
Since this point lies on the line with equation $x+y=1$, then $\frac{1}{2} q+\frac{1}{2} q=1$ or $q=1$.
Thus, $Q$ has coordinates $(1,1)$.
Therefore, $\ell^{\prime}$ passes through $Q(1,1)$ and $P\left(\frac{1}{k+1}, \frac{k}{k+1}\right)$.
The slope of $\ell^{\prime}$ is thus $\frac{1-\frac{k}{k+1}}{1-\frac{1}{k+1}}=\frac{\frac{k+1}{k+1}-\frac{k}{k+1}}{\frac{k+1}{k+1}-\frac{1}{k+1}}=\frac{\frac{1}{k+1}}{\frac{k}{k+1}}=\frac{1}{k}$. (Note that $k \neq 0$.)
Therefore, $\ell^{\prime}$ has equation $y=\frac{1}{k} x+b$ for some $b$.
Since $Q(1,1)$ lies on $\ell^{\prime}$, then $1=\frac{1}{k}+b$ or $b=\frac{k-1}{k}$.
Therefore, the slope of $\ell^{\prime}$ is $\frac{1}{k}$ and its $y$-intercept is $\frac{k-1}{k}$.
ANSWER: Slope is $\frac{1}{k} ; y$-intercept is $\frac{k-1}{k}$