CAT 2019 Shift 2 Paper

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Question : 98 of 100

Marks: +1, -0

Let a, b, x, y be real numbers such that

a^2+b^2=25,x^2+y^2=169,

and

ax+by=65 .

if

k=ay-bx,

then

$k=5/13$
$0 < k ≤ 5/13$
$k=0$
$k > 5/13$

Solution:

We can take

a=5, b=0, x=13

and

y=0

as values which satisfies all three equations.
Hence,

k=a y-b x=5 × 0-0 × 13=0

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