Question

\(D,E,F\) are respectively the points on the sides \(BC,CA\) and \(AB\) of a \(\triangle ABC\), dividing them in the ratio \(2:3,\;1:2,\;3:1\) internally. The lines \(BE\) and \(CF\) intersect on the line \(AD\) at \(P\). If \[ \overrightarrow{AP}=x_1\overrightarrow{AB}+y_1\overrightarrow{AC}, \] then \[ x_1+y_1= \]

• A. \(\dfrac{5}{6}\)
• B. \(1\)
• C. \(\dfrac{3}{2}\)
• D. \(2\)

Hint:

For vector geometry problems, take one vertex as origin and express all points as linear combinations of two independent vectors such as \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).

Answer 1

The Correct Option is A

Step 1: Choose a convenient origin.
Take $A$ as the origin and let $\overrightarrow{AB}=\vec b$, $\overrightarrow{AC}=\vec c$. So $B=\vec b$ and $C=\vec c$.
Step 2: Locate $D$, $E$, $F$.
$D$ divides $BC$ in $2:3$, so $D=\dfrac{3\vec b+2\vec c}{5}$. $E$ divides $CA$ in $1:2$, so $E=\dfrac{2}{3}\vec c$. $F$ divides $AB$ in $3:1$, so $F=\dfrac{3}{4}\vec b$.
Step 3: Write $P$ on line $AD$.
Any point on $AD$ is $P=t\,D=t\left(\dfrac{3\vec b+2\vec c}{5}\right)=\dfrac{3t}{5}\vec b+\dfrac{2t}{5}\vec c$.
Step 4: Write $P$ on line $CF$.
A point on $CF$ is $P=(1-s)\vec c+s\cdot\dfrac{3}{4}\vec b=\dfrac{3s}{4}\vec b+(1-s)\vec c$. Matching coefficients with Step 3: $\dfrac{3t}{5}=\dfrac{3s}{4}$ and $\dfrac{2t}{5}=1-s$. From the first, $s=\dfrac{4t}{5}$; substituting, $\dfrac{2t}{5}=1-\dfrac{4t}{5}$, giving $\dfrac{6t}{5}=1$, so $t=\dfrac{5}{6}$.
Step 5: Get the coordinates of $P$.
Then $\overrightarrow{AP}=\dfrac{3t}{5}\vec b+\dfrac{2t}{5}\vec c=\dfrac{1}{2}\vec b+\dfrac{1}{3}\vec c$. So $x_1=\dfrac{1}{2}$ and $y_1=\dfrac{1}{3}$.
Step 6: Add the coefficients.
Hence $x_1+y_1=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$.
\[ \boxed{\dfrac{5}{6}} \]