For any two points M and N in the XY-plane, let $\overrightarrow{\mathrm{MN}}$ denote the vector from M to N, and $\overrightarrow{0}$ denote the zero vector.

Let P, Q and R be three distinct points in the XY-plane. Let S be a point inside the triangle ΔPQR such that

\overrightarrow{\mathrm{SP}}+5\overrightarrow{\mathrm{SQ}}+6\overrightarrow{\mathrm{SR}}=\overrightarrow{0}

Let E and F be the mid-points of the sides PR and QR, respectively. Then the value of

length of the line segment EF \text{/length of the line segment}\mathrm{ES}

is ______.

The correct answer is 1.20.

Let the position vectors of the points $P$, $Q$, $R$, and $S$ relative to some origin be denoted by $\overrightarrow{p}$, $\overrightarrow{q}$, $\overrightarrow{r}$, and $\overrightarrow{s}$, respectively.
The given vector relation is:
$\overrightarrow{\mathrm{SP}}+5\overrightarrow{\mathrm{SQ}}+6\overrightarrow{\mathrm{SR}}=\overrightarrow{0}$

We can express the vectors in terms of their position vectors:
$(\overrightarrow{p}-\overrightarrow{s})+5(\overrightarrow{q}-\overrightarrow{s})+6(\overrightarrow{r}-\overrightarrow{s})=\overrightarrow{0}$
Simplifying this expression gives:
$\overrightarrow{p}+5\overrightarrow{q}+6\overrightarrow{r}=12\overrightarrow{s}$
Therefore, the position vector of $S$ is:
$\overrightarrow{s}=\frac{\overrightarrow{p}+5\overrightarrow{q}+6\overrightarrow{r}}{12}$

We are given that $E$ is the mid-point of the side $\mathrm{PR}$. Therefore, the position vector of $E$, denoted by $\overrightarrow{e}$, is:
$\overrightarrow{e}=\frac{\overrightarrow{p}+\overrightarrow{r}}{2}$
Similarly, $F$ is the mid-point of $\mathrm{QR}$. The position vector of $F$, denoted by $\overrightarrow{f}$, is:
$\overrightarrow{f}=\frac{\overrightarrow{q}+\overrightarrow{r}}{2}$

Now, let us find the vector $\overrightarrow{\mathrm{EF}}$ representing the line segment $\mathrm{EF}$:
$\overrightarrow{\mathrm{EF}}=\overrightarrow{f}-\overrightarrow{e}=\frac{\overrightarrow{q}+\overrightarrow{r}}{2}-\frac{\overrightarrow{p}+\overrightarrow{r}}{2}=\frac{\overrightarrow{q}-\overrightarrow{p}}{2}$
Thus, the length of the line segment $\mathrm{EF}$ is:
$\text{length of}\mathrm{EF}=\left|\overrightarrow{\mathrm{EF}}\right|=\frac{1}{2}|\overrightarrow{q}-\overrightarrow{p}|$

Next, let us determine the vector $\overrightarrow{\mathrm{ES}}$ representing the line segment $\mathrm{ES}$:
$\overrightarrow{\mathrm{ES}}=\overrightarrow{s}-\overrightarrow{e}=\frac{\overrightarrow{p}+5\overrightarrow{q}+6\overrightarrow{r}}{12}-\frac{\overrightarrow{p}+\overrightarrow{r}}{2}$
Expressing both terms over a common denominator of 12:
$\overrightarrow{\mathrm{ES}}=\frac{\overrightarrow{p}+5\overrightarrow{q}+6\overrightarrow{r}-6(\overrightarrow{p}+\overrightarrow{r})}{12}$
$\overrightarrow{\mathrm{ES}}=\frac{\overrightarrow{p}+5\overrightarrow{q}+6\overrightarrow{r}-6\overrightarrow{p}-6\overrightarrow{r}}{12}=\frac{5\overrightarrow{q}-5\overrightarrow{p}}{12}=\frac{5}{12}(\overrightarrow{q}-\overrightarrow{p})$
Thus, the length of the line segment $\mathrm{ES}$ is:
$\text{length of}\mathrm{ES}=\left|\overrightarrow{\mathrm{ES}}\right|=\frac{5}{12}|\overrightarrow{q}-\overrightarrow{p}|$

We are asked to find the value of the ratio:
$\frac{\text{length of}\mathrm{EF}}{\text{length of}\mathrm{ES}}=\frac{\frac{1}{2}|\overrightarrow{q}-\overrightarrow{p}|}{\frac{5}{12}|\overrightarrow{q}-\overrightarrow{p}|}=\frac{1/2}{5/12}=\frac{12}{10}=1.20$