SAT Question adae6543: Lines, Angles, and Triangles answer and explanation with Desmos solution

Choice D is correct. The figure shows that angle $MRL$ and angle $PRQ$ are vertical angles. Since vertical angles are congruent, angle $MRL$ and angle $PRQ$ are congruent. It's given that $\overline{LM}$ is parallel to $\overline{PQ}$. The figure also shows that $\overline{LQ}$ intersects $\overline{LM}$ and $\overline{PQ}$. If two parallel segments are intersected by a third segment, alternate interior angles are congruent. Thus, alternate interior angles $MLR$ and $PQR$ are congruent. Since triangles $LMR$ and $PQR$ have two pairs of congruent angles, the triangles are similar. Sides $LR$ and $MR$ in triangle $LMR$ correspond to sides $RQ$ and $RP$, respectively, in triangle $PQR$. Since the lengths of corresponding sides in similar triangles are proportional, it follows that $\frac{RQ}{LR}=\frac{RP}{MR}$. It's given that the lengths of $\overline{MR}$, $\overline{LR}$, and $\overline{RP}$ are $6$, $7$, and $11$, respectively. Substituting $6$ for $MR$, $7$ for $LR$, and $11$ for $RP$ in the equation $\frac{RQ}{LR}=\frac{RP}{MR}$ yields $\frac{RQ}{7}=\frac{11}{6}$. Multiplying each side of this equation by $7$ yields $RQ=\left(\frac{11}{6}\right)\left(7\right)$, or $RQ=\frac{77}{6}$. It's given that $\overline{LQ}$ intersects $\overline{MP}$ at point $R$, so $LQ=LR+RQ$. Substituting $7$ for $LR$ and $\frac{77}{6}$ for $RQ$ in this equation yields $LQ=7+\frac{77}{6}$, or $LQ=\frac{119}{6}$. Therefore, the length of $\overline{LQ}$ is $\frac{119}{6}$.