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KEAM 2013 Math Paper

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Question : 55 of 120

Marks: +1, -0

If the equation

3x + 3y + 5 = 0

is written in the form x

\cos\alpha + y \sin\alpha = p,

then the value of

\sin\alpha + \cos\alpha

is

$\sqrt{2}$
$\frac{1}{\sqrt{2}}$
$-\sqrt{2}$
$-\frac{1}{\sqrt{2}}$
$\sqrt{3}$

Solution:

Given equation of straight line is

3x + 3y + 5 = 0

\Rightarrow \;\; 3x + 3y = -5

\Rightarrow \;\; -3x - 3y = 5

Now, for converting this equation in normal form.

Divide by

\sqrt{(-3)^2 + (-3)^2} = 3\sqrt{2}

on both sides, we get

\;\frac{-3}{3\sqrt{2}}x - \;\frac{3}{3\sqrt{2}}y = \;\frac{5}{3\sqrt{2}}

\Rightarrow \;\; -\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y = \frac{5}{3\sqrt{2}}

Compare with

x \cos\alpha + y \sin\alpha = p,

we get

\cos\alpha = \sin\alpha = -\frac{1}{\sqrt{2}}

\therefore \sin\alpha + \cos\alpha = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{-2}{\sqrt{2}} = -\sqrt{2}

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