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: Place coordinates and find P, Q, R, S.
\(A=(0,0)\), \(B=(0,6)\), \(C=(9,0)\). P and Q lie on AC with \(AP=PQ=QC\), so they split AC into three equal parts:
\[ P = (3,0), \quad Q = (6,0) \]
R and S lie on AB with \(AR=RS=SB\), splitting AB into three equal parts:
\[ R = (0,2), \quad S = (0,4) \]

Step 2: Write the equation of line PB.
PB joins \(P=(3,0)\) and \(B=(0,6)\). Its slope is:
\[ \frac{6-0}{0-3} = -2 \]
So the line is \(y - 0 = -2(x-3)\), i.e.
\[ y = -2x + 6 \]

Step 3: Write the equation of line RC.
RC joins \(R=(0,2)\) and \(C=(9,0)\). Its slope is:
\[ \frac{0-2}{9-0} = -\frac{2}{9} \]
So the line is \(y - 2 = -\frac{2}{9}(x-0)\), i.e.
\[ y = 2 - \frac{2}{9}x \]

Step 4: Solve the two equations together to find X.
\[ -2x + 6 = 2 - \frac{2}{9}x \]
\[ -2x + \frac{2}{9}x = 2 - 6 \]
\[ -\frac{16}{9}x = -4 \]
\[ x = \frac{9}{4} \]
Put back: \(y = -2\left(\frac{9}{4}\right) + 6 = -\frac{9}{2}+6 = \frac{3}{2}\). So \(X = \left(\frac94, \frac32\right)\).

Step 5: Find the slope of AX.
\(A = (0,0)\), so the slope of AX is simply \( \frac{y_X}{x_X} \):
\[ \text{slope} = \frac{3/2}{9/4} = \frac{3}{2}\times\frac{4}{9} = \frac{12}{18} = \frac{2}{3} \]
This matches option 1, not \(-\frac23\), \(\frac32\) or \(-\frac34\), since the line AX rises to the right through the first quadrant with this exact slope.

Final Answer:
\[ \boxed{\text{slope of } AX = \frac{2}{3}} \]