Answer: C. $90^{\circ}-t^{\circ}$

Explanation:1. Analyze Triangle PQR: Vertices are $P(2,5)$, $Q(2,9)$, $R(6,5)$. Side PQ is vertical because the x-coordinates are the same ($2$). Length of PQ = $|9-5| = 4$. Side PR is horizontal because the y-coordinates are the same ($5$). Length of PR = $|6-2| = 4$. Since PQ is vertical and PR is horizontal, $\angle P$ is a right angle ($90^{\circ}$). Therefore, Triangle PQR is a right triangle. The sum of angles in a triangle is $180^{\circ}$. So, $\angle P + \angle Q + \angle R = 180^{\circ}$. $90^{\circ} + t^{\circ} + \angle R = 180^{\circ}$ $\angle R = 180^{\circ} - 90^{\circ} - t^{\circ} = 90^{\circ} - t^{\circ}$.

2. Analyze Triangle LMN: Vertices are $L(2,5)$, $M(2,9+k)$, $N(6+k,5)$. Side LM is vertical because the x-coordinates are the same ($2$). Length of LM = $|(9+k)-5| = |4+k|$. Since k is positive, $4+k$ is positive. So, length is $4+k$. Side LN is horizontal because the y-coordinates are the same ($5$). Length of LN = $|(6+k)-2| = |4+k|$. Since k is positive, $4+k$ is positive. So, length is $4+k$. Since LM is vertical and LN is horizontal, $\angle L$ is a right angle ($90^{\circ}$). Therefore, Triangle LMN is a right triangle. Also, since LM = LN = $4+k$, Triangle LMN is an isosceles right triangle. This means $\angle M = \angle N = 45^{\circ}$. However, this contradicts the relationship with $\angle Q = t^{\circ}$ unless $t=45^{\circ}$. Let's not assume it's isosceles from this step yet, but focus on it being a right triangle.

3. Relate the angles in Triangle LMN: Since Triangle LMN is a right triangle with $\angle L = 90^{\circ}$: $\angle L + \angle M + \angle N = 180^{\circ}$ $90^{\circ} + \angle M + \angle N = 180^{\circ}$ $\angle M + \angle N = 90^{\circ}$

4. Consider similarity or specific properties: In Triangle PQR, side lengths are PQ=$4$, PR=$4$. This means PQR is an isosceles right triangle, so $\angle Q = \angle R = 45^{\circ}$. Therefore $t=45^{\circ}$. Then $\angle N$ in LMN should also be $45^{\circ}$ because LMN is also an isosceles right triangle (LM=LN=$4+k$). If $t=45^{\circ}$, then $90^{\circ}-t^{\circ} = 90^{\circ}-45^{\circ} = 45^{\circ}$. This matches option [C].

Let's verify without assuming PQR is isosceles initially, only that it's a right triangle at P.In right $\triangle PQR$, $\angle P = 90^{\circ}$, $\angle Q = t^{\circ}$, so $\angle R = 90^{\circ} - t^{\circ}$.In right $\triangle LMN$, $\angle L = 90^{\circ}$. We need $\angle N$.The transformation from PQR to LMN:P(2,5) maps to L(2,5) (same point).Q(2,9) maps to M(2,9+k).R(6,5) maps to N(6+k,5).This is not a simple similarity transformation unless $k=0$.However, $\triangle PQR$ has legs PQ (vertical, length $|9-5|=4$) and PR (horizontal, length $|6-2|=4$).$\triangle LMN$ has legs LM (vertical, length $|(9+k)-5|=|4+k|$) and LN (horizontal, length $|(6+k)-2|=|4+k|$).Since both triangles are right-angled at P and L respectively, and both have equal leg lengths (PQ=PR and LM=LN), they are both isosceles right triangles.Therefore, for $\triangle PQR$, $\angle Q = \angle R = 45^{\circ}$. So, $t=45^{\circ}$.For $\triangle LMN$, $\angle M = \angle N = 45^{\circ}$.Thus, $\angle N = 45^{\circ}$.Now check the options if $t=45^{\circ}$:[A] $(90-(45-k))^{\circ} = (45+k)^{\circ}$[B] $(90-(45+k))^{\circ} = (45-k)^{\circ}$[C] $90^{\circ}-45^{\circ} = 45^{\circ}$[D] $90^{\circ}-k^{\circ}$Option [C] gives $45^{\circ}$.

The question implies a general relationship for any positive constant k and any $t^{\circ}$ that forms a valid angle Q in a right triangle PQR.The specific coordinates for PQR reveal it IS an isosceles right triangle, making $t=45^{\circ}$.If PQR was not isosceles, say R was $(x\_R, 5)$ with $x\_R \neq 6$.Then PR length would be $|x\_R-2|$. $\tan(t) = \dfrac{PR}{PQ} = \dfrac{|x\_R-2|}{4}$.And in LMN, LN length would be $|x\_R+k-2|$. $\tan(\angle M) = \dfrac{LN}{LM} = \dfrac{|x\_R+k-2|}{|4+k|}$.The problem likely relies on the specific geometry given. Both triangles are right-angled at P and L respectively.In any right triangle LMN with $\angle L = 90^{\circ}$, we have $\angle M + \angle N = 90^{\circ}$, so $\angle N = 90^{\circ} - \angle M$.The vertices define the triangles uniquely.Triangle PQR is a right triangle at P ($90^{\circ}$). $\angle Q = t^{\circ}$. Then $\angle R = 90^{\circ} - t^{\circ}$.Triangle LMN is a right triangle at L ($90^{\circ}$).This problem is about the relationship between angles in these specific triangles.Since both triangles are right-angled (at P and L), and based on the coordinates, their legs are along grid lines.PQR: Legs PQ=$4$, PR=$4$. So $t=45^{\circ}$. $\angle R = 45^{\circ}$.LMN: Legs LM=$4+k$, LN=$4+k$. So $\angle M = \angle N = 45^{\circ}$.The question seems simpler than a general proof of similarity if $t$ can vary.If $\angle Q = t^{\circ}$, and we found $t=45^{\circ}$, then $\angle N = 45^{\circ}$.The answer choice $90^{\circ}-t^{\circ}$ would be $90^{\circ}-45^{\circ}=45^{\circ}$. This holds.

This implies that $\angle N$ is always equal to $\angle R$ from $\triangle PQR$.Is $\triangle PQR \sim \triangle LMN$?Ratio of legs in PQR is $PQ/PR = 4/4 = 1$.Ratio of legs in LMN is $LM/LN = (4+k)/(4+k) = 1$.Yes, both are isosceles right triangles. All isosceles right triangles are similar to each other.Therefore, their corresponding angles are equal.$\angle P = \angle L = 90^{\circ}$.$\angle Q$ corresponds to $\angle M$. So $\angle M = t^{\circ}$.$\angle R$ corresponds to $\angle N$. So $\angle N = \angle R$.Since $\angle R = 90^{\circ} - t^{\circ}$ (from $\triangle PQR$), then $\angle N = 90^{\circ} - t^{\circ}$.

Conclusion:The measure of $\angle N$ is $90^{\circ}-t^{\circ}$.