Let $t$ be TNYWR.

The triangle in the first quadrant formed by the $x$-axis, the $y$-axis, and the line with equation $(k-1) x+(k+1) y=t$ has area 10. What is the value of $k$ ?
To find the $x$-intercept of the line with equation $(k-1) x+(k+1) y=t$, we set $y=0$ and obtain $(k-1) x=t$ or $x=\frac{t}{k-1}$. We assume that $k \neq 1$.

To find the $y$-intercept of the line with equation $(k-1) x+(k+1) y=t$, we set $x=0$ and obtain $(k+1) y=t$ or $y=\frac{t}{k+1}$. We assume that $k \neq-1$.

The triangle formed by the $x$-axis, the $y$-axis, and this line has vertices at the $x$-intercept, the $y$-intercept, and the origin.

This triangle is right-angled at the origin, so its area equals $\frac{1}{2} \cdot \frac{t}{k-1} \cdot \frac{t}{k+1}=\frac{t^{2}}{2\left(k^{2}-1\right)}$. Since we are told that this area is 10 , then $\frac{t^{2}}{2\left(k^{2}-1\right)}=10$ which gives $k^{2}-1=\frac{t^{2}}{20}$ or $k^{2}=1+\frac{t^{2}}{20}$.

Since the answer to (b) is 5 , then $t=5$.

Therefore, $k^{2}=1+\frac{5^{2}}{20}=1+\frac{5}{4}=\frac{9}{4}$.

Since $t>0$, then for the $x$ - and $y$-intercepts of the line to be positive (putting the triangle in the first quadrant), then $k>0$.

Since $k^{2}=\frac{9}{4}$, then $k=\frac{3}{2}$.

ANSWER: $11,5, \frac{3}{2}$