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

Approach: The one fact that unlocks a parallelogram is that its diagonals bisect each other. So the midpoint of \(PR\) equals the midpoint of \(SQ\) — that single condition pins down \(S\), and then \(SQ\) is just a line to intersect with the x-axis.

Step 1: Midpoint of diagonal \(PR\).
\(P(-3,-2),\ R(9,1)\):
\[ M = \left(\frac{-3+9}{2}, \frac{-2+1}{2}\right) = \left(3, -\tfrac12\right). \]

Step 2: Find \(S\) from the same midpoint of \(SQ\).
Let \(S = (x, y)\), with \(Q(1, -5)\). The midpoint of \(SQ\) is also \((3, -\tfrac12)\):
\[ \frac{x+1}{2} = 3 \Rightarrow x = 5, \qquad \frac{y-5}{2} = -\tfrac12 \Rightarrow y = 4. \]
So \(S = (5, 4)\).

Step 3: Equation of line \(SQ\).
Through \(S(5,4)\) and \(Q(1,-5)\):
\[ \text{slope} = \frac{-5-4}{1-5} = \frac{-9}{-4} = \frac94. \]
\[ y - 4 = \frac94(x - 5) \Rightarrow 4y - 16 = 9x - 45 \Rightarrow 9x - 4y - 29 = 0. \]

Step 4: Cross the x-axis.
Set \(y = 0\): \(9a - 29 = 0 \Rightarrow a = \frac{29}{9}.\)

\(a = \dfrac{29}{9}\) — option (1).

Answer 2

Approach: Since the diagonals of a parallelogram bisect each other, the midpoint of \(PR\) is the SAME point as the midpoint of \(SQ\). That means this midpoint already lies on diagonal \(SQ\), so the line \(SQ\) is simply the line through \(Q\) and this midpoint. There is no need to solve for \(S\) at all.

Step 1: Midpoint of \(PR\).
\(P(-3,-2)\), \(R(9,1)\):
\[ M = \left(\frac{-3+9}{2}, \frac{-2+1}{2}\right) = \left(3, -\frac{1}{2}\right). \]
This \(M\) is also the midpoint of \(SQ\), so \(M\) lies on line \(SQ\).

Step 2: Line through \(Q\) and \(M\).
\(Q(1,-5)\), \(M\left(3, -\frac{1}{2}\right)\): slope \( = \dfrac{-\frac{1}{2} - (-5)}{3 - 1} = \dfrac{\frac{9}{2}}{2} = \frac{9}{4}\).
\[ y + 5 = \frac{9}{4}(x - 1). \]

Step 3: Intersect the x-axis.
Set \(y = 0\): \(5 = \frac{9}{4}(x-1) \Rightarrow x - 1 = \frac{20}{9} \Rightarrow x = \frac{29}{9}\).

\(a = \dfrac{29}{9}\), option (1).