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: First locate \(S\) using the shared midpoint: \(S = P + R - Q = (-3,-2) + (9,1) - (1,-5) = (5, 4)\).
Step 2: The x-axis point \((a, 0)\) must be collinear with \(Q(1,-5)\) and \(S(5,4)\). For three points \((x_1,y_1),(x_2,y_2),(x_3,y_3)\) to be collinear, \(x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0\). Using \((a,0)\), \(Q(1,-5)\), \(S(5,4)\):
\[ a(-5 - 4) + 1(4 - 0) + 5(0 - (-5)) = 0 \Rightarrow -9a + 4 + 25 = 0. \]
Step 3: Solve: \(-9a + 29 = 0 \Rightarrow a = \dfrac{29}{9}\).
\[ \boxed{a = \frac{29}{9}} \]