Question

Triangle ABC has vertices A(0, 0), B(0, 6) and C(9, 0). The points P and Q lie on side AC such that AP = PQ = QC. Similarly, the points R and S lie on side AB such that AR = RS = SB. If the line segments PB and RC intersect at X, then the slope of the line AX is:

• A. \(\frac{2}{3}\)
• B. \(-\frac{2}{3}\)
• C. \(\frac{3}{2}\)
• D. \(-\frac{3}{4}\)

Hint:

Find P, Q, R, S by dividing AC and AB into three equal parts, write the equations of lines PB and RC, solve them together for X, then compute the slope of AX.

Answer 1

The Correct Option is A

Step 1: Locate the key points.
$A=(0,0)$, $B=(0,6)$, $C=(9,0)$. Since $AP=PQ=QC$ splits $AC$ into three equal parts, $P=(3,0)$ and $Q=(6,0)$. Since $AR=RS=SB$ splits $AB$ into three equal parts, $R=(0,2)$ and $S=(0,4)$.

Step 2: Write PB and RC as parametric lines.
Any point on segment PB can be written as a mix of P and B using a parameter $t$:
\[ (x,y) = P + t(B-P) = (3,0) + t(-3, 6) = (3-3t,\ 6t) \]
Any point on segment RC can be written using a parameter $s$:
\[ (x,y) = R + s(C-R) = (0,2) + s(9,-2) = (9s,\ 2-2s) \]

Step 3: Match the two parametric forms at X.
Equating x-coordinates and y-coordinates:
\[ 3-3t = 9s \]
\[ 6t = 2-2s \]
From the second equation, $t = \frac{2-2s}{6} = \frac{1-s}{3}$. Substitute into the first:
\[ 3 - 3\left(\frac{1-s}{3}\right) = 9s \]
\[ 3 - (1-s) = 9s \]
\[ 2+s = 9s \]
\[ 8s = 2 \implies s = \frac{1}{4} \]

Step 4: Get X from s.
\[ X = (9s,\ 2-2s) = \left(9\times\frac14,\ 2-2\times\frac14\right) = \left(\frac94,\ \frac32\right) \]
This point lands on segment RC (since $0 \le s=\frac14 \le 1$), and using $t=\frac{1-s}{3}=\frac14$ in the PB equation gives the same point $(\frac94,\frac32)$, so it also lies on PB, confirming X is correct.

Step 5: Compute the slope of AX.
\[ \text{slope of } AX = \frac{3/2 - 0}{9/4 - 0} = \frac{3/2}{9/4} = \frac23 \]

Final Answer:
\[ \boxed{\dfrac{2}{3}} \]