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UPSC NDA I 2016 Mathematics Solved Paper

For the next three (3) items that follow :

Consider a parallelogram whose vertices are

A(1,2),B(4,y),C(x,6)

and

D(3,5)

taken in order.

[NDA I 2016]

Question : 58 of 120

Marks: +1, -0

What is the value of

AC^2 - BD^2

Solution:

Suppose Mid point of AC and BD is M (a, b).

a = \;\;\frac{1+x}{2} = \;\;\frac{3+4}{2}

\Rightarrow x=6

b = \;\;\frac{5+y}{2} = \;\;\frac{2+6}{2}

\Rightarrow y=3

a = \;\;\frac{7}{2}, b=4

AC^{2}=(1-x)^{2}+(2-6)^{2}=(1-6)^{2}+(-4)^{2}=41

BD^{2}=(3-4)^{2}+(5-3)^{2}=1+4=5

AC^{2}-BD^{2}=41-5

AC^{2}-BD^{2}=36

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