Question

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

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $-\sqrt{2}$
• D. $-\frac{1}{\sqrt{2}}$
• E. $\sqrt{3}$

Answer 1

Answer: (C)

$-\sqrt{2}$

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