Question

ABCD is a quadrilateral with \(A(3,5)\), \(B(5,9)\) and \(C(3,7)\). If the line \(y=3x\) meets the sides \(AB\) and \(BC\) in points \(P\) and \(Q\) respectively, then \(PQ^2=\)

• A. \(121\)
• B. \(90\)
• C. \(169\)
• D. \(150\)

Hint:

To find the intersection of two lines:
• Find the equation of each line.
• Solve them simultaneously.
• Use the distance formula: \[ PQ^2=(x_2-x_1)^2+(y_2-y_1)^2. \]

Answer 1

The Correct Option is B

Step 1: Find the coordinates of \(P\). Equation of line \(AB\): \[ m=\frac{9-5}{5-3}=2 \] \[ y-5=2(x-3) \] \[ y=2x-1 \] Since \(P\) lies on \(y=3x\), \[ 3x=2x-1 \] \[ x=-1,\qquad y=-3 \] Hence, \[ P=(-1,-3) \]

Step 2: Find the coordinates of \(Q\). Equation of line \(BC\): \[ m=\frac{7-9}{3-5}=1 \] \[ y-9=x-5 \] \[ y=x+4 \] Since \(Q\) lies on \(y=3x\), \[ 3x=x+4 \] \[ 2x=4 \] \[ x=2,\qquad y=6 \] Hence, \[ Q=(2,6) \]

Step 3: Calculate \(PQ^2\). \[ PQ^2=(2+1)^2+(6+3)^2 \] \[ =3^2+9^2 \] \[ =9+81=90 \]

Step 4: Final conclusion. \[ \boxed{PQ^2=90} \] Hence, the correct option is \(\boxed{(B)}\).