Question

If the point $(3,4)$ lies on the locus of the point of intersection of the lines $x\cos \alpha +y\sin \alpha =a$ and $x\sin \alpha -y\cos \alpha =b$ where $\alpha$ is a variable), the point $(a,b)$ lies on line $3x-4y=0,$ then $|a+b|$ is equal to

• A. $1$
• B. $7$
• C. $12$
• D. $5$

Answer 1

Answer: (B)

$7$

Squaring and adding the given equations of the lines we get $x^2+y^2=a^2+b^2$ as the locus of the point of intersection of these lines.
Since $(3,4)$ lies on the locus, we get
$9+16=a^2+b^2\Rightarrow a^2+b^2=25$ ...(1)
Also $(a,b)$ lies on $3x-4y=0$
So $3a-4b=0\Rightarrow b=\left(\frac{3}{4}\right)a$ ...(2)
From (1) $a^2+\left(\frac{9}{16}\right)a^2=25\Rightarrow a^2=16$
so that ${|a+b|}^2={\left(\frac{7}{4}\right)}^2{a}^2=49$
$\therefore |a+b|=7$