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UPSC NDA I 2016 Mathematics Solved Paper

Question Numbers: 58-60

Consider a parallelogram whose vertices are A (1, 2), B (4, y), C (x, 6) and D (3, 5) taken in order

Question : 58 of 120

Marks: +1, -0

AC^{2} - BD^{2}

Solution:

Concept:

In a parallelogram, the diagonals bisect each other. The midpoint of diagonal

AC

equals the midpoint of diagonal

BD

Explanation:

Given vertices

A(1,2)

B(4,y)

C(x,6)

D(3,5)

Using the midpoint formula:

Midpoint of

AC = \left( \frac{1+x}{2}, \frac{2+6}{2} \right) = \left( \frac{1+x}{2}, 4 \right)

Midpoint of

BD = \left( \frac{4+3}{2}, \frac{y+5}{2} \right) = \left( \frac{7}{2}, \frac{y+5}{2} \right)

Equate:

\frac{1+x}{2} = \frac{7}{2}

and

4 = \frac{y+5}{2}

Solving:

1+x = 7 \Rightarrow x = 6

;

8 = y+5 \Rightarrow y = 3

Now

B(4,3)

C(6,6)

Compute

AC^2

AC = \sqrt{(6-1)^2 + (6-2)^2} = \sqrt{25 + 16} = \sqrt{41}

, so

AC^2 = 41

Compute

BD^2

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

, so

BD^2 = 5

Therefore

AC^2 - BD^2 = 41 - 5 = 36

Answer:

36

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