Question

The $(x, y)$ coordinates of vertices $P$, $Q$ and $R$ of a parallelogram $PQRS$ are $(-3, -2)$, $(1, -5)$ and $(9, 1)$, respectively. If the diagonal $SQ$ intersects the x-axis at $(a, 0)$, then the value of $a$ is:

• A. \(\frac{29}{9}\)
• B. \(\frac{27}{9}\)
• C. \(\frac{29}{7}\)
• D. \(\frac{9}{29}\)

Hint:

In any parallelogram, diagonals always bisect each other. So, to find missing vertices, equate midpoints of the diagonals. This avoids unnecessary vector calculations and simplifies coordinate geometry problems.

Answer 1

The Correct Option is A

Step 1: Use the property of diagonals in a parallelogram. In any parallelogram, diagonals bisect each other. Thus, the midpoint of diagonal \(PR\) is the same as the midpoint of diagonal \(SQ\).
Step 2: Find the midpoint of diagonal \(PR\). Coordinates: \[ P(-3, -2),\quad R(9, 1). \] Midpoint: \[ M = \left( \frac{-3+9}{2},\, \frac{-2+1}{2} \right) = (3, -\tfrac{1}{2}). \]
Step 3: Find coordinates of point \(S\). Let \(S = (x, y)\). Given midpoint of \(SQ\) is also \((3, -\tfrac12)\), and \(Q = (1, -5)\). Using midpoint formula: \[ \frac{x + 1}{2} = 3 \quad\Rightarrow\quad x + 1 = 6 \Rightarrow x = 5, \] \[ \frac{y - 5}{2} = -\tfrac12 \quad\Rightarrow\quad y - 5 = -1 \Rightarrow y = 4. \] Thus, \[ S = (5, 4). \]
Step 4: Equation of diagonal \(SQ\). Points on line: \(S(5,4)\) and \(Q(1,-5)\). Slope: \[ m = \frac{-5 - 4}{1 - 5} = \frac{-9}{-4} = \frac{9}{4}. \] Using point-slope form at \(S(5,4)\): \[ y - 4 = \frac{9}{4}(x - 5). \] Multiply through: \[ 4y - 16 = 9x - 45, \] \[ 9x - 4y - 29 = 0. \]
Step 5: Find intersection with x-axis. At the x-axis, \(y = 0\). Substitute into the line equation: \[ 9a - 0 - 29 = 0 \Rightarrow 9a = 29 \Rightarrow a = \frac{29}{9}. \] Thus, \[ \boxed{\frac{29}{9}}. \]