Q208 / 270 · NDA 2016 (Phase 1) · Mathematics
Consider a parallelogram whose vertices are $\mathrm{A}(1,2), \mathrm{B}(4$, y ), $\mathrm{C}(\mathrm{x}, 6)$ and $\mathrm{D}(3,5)$ taken in order.
What is the value of $\mathrm{AC}^2-\mathrm{BD}^2$ ?
- A25
- B30
- Correct answerC36
- D40
Solution
Suppose Mid point of $A C$ and $B D$ is $M(a, b)$.
$\begin{aligned} & a=\frac{1+x}{2}=\frac{3+4}{2} \\ & \Rightarrow x=6 \\ & b=\frac{5+y}{2}=\frac{2+6}{2}\end{aligned}$ $\begin{aligned} & \Rightarrow y=3 \\ & a=\frac{7}{2}, b=4 \\ & A C^2=(1-x)^2+(2-6)^2=(1-6)^2+(-4)^2=41 \\ & B D^2=(3-4)^2+(5-3)^2=1+4=5 \\ & A C^2-B D^2=41-5 \\ & A C^2-B D^2=36\end{aligned}$
$\begin{aligned} & a=\frac{1+x}{2}=\frac{3+4}{2} \\ & \Rightarrow x=6 \\ & b=\frac{5+y}{2}=\frac{2+6}{2}\end{aligned}$ $\begin{aligned} & \Rightarrow y=3 \\ & a=\frac{7}{2}, b=4 \\ & A C^2=(1-x)^2+(2-6)^2=(1-6)^2+(-4)^2=41 \\ & B D^2=(3-4)^2+(5-3)^2=1+4=5 \\ & A C^2-B D^2=41-5 \\ & A C^2-B D^2=36\end{aligned}$
Hence, the correct answer is option: 'C'.