Question

Let $P(a,b)$ and $Q(c,d)$ be the two points on the line $3x-4y=1$ which are at a distance of $5\text{units}$ from the point $(3,2)$, and $a>0,b>0,c<0,d<0$. Then which of the following is/are true

• A. distance from origin to $P$ is $\sqrt{2}$ units
• B. $a+b+c+d=10$
• C. quadratic equation whoe roots are $2c,2d$ is $x^2+4x+2=0$
• D. equation of locus of a point equidistant from $P$ and $Q$ is $4x+3y=18$

Answer 1

Answer: (B), (D)

equation of locus of a point equidistant from $P$ and $Q$ is $4x+3y=18$

Given line : $3x-4y=1$
$\therefore m=\tan \theta =\frac{3}{4}\Rightarrow \sin \theta =\frac{3}{5},\cos \theta =\frac{4}{5},r=5$
$\Rightarrow$required points $\equiv (3\pm 5\times \frac{4}{5},2\pm 5\times \frac{3}{5})$
$\equiv (3\pm 4,2\pm 3)$
$\Rightarrow P\equiv (7,5)$ and $Q\equiv (-1,-1)$
$\Rightarrow OP=\sqrt{7^2+5^2}=\sqrt{74},$
$a+b+c+d=7+5+(-1)+(-1)=10$
Now, quadratic equation whose roots are $2c,2d=-2$ is : $x^2-(-4)x+4=0\Rightarrow x^2+4x+4=0$

Locus of a point equidistant from $P$ and $Q$ is perpendicular bisector of $PQ.$
Mid point of $PQ\equiv (\frac{7-1}{2},\frac{5-1}{2})\equiv (3,2)$
$\therefore$equation of pependicular bisector is : $y-2=-\frac{4}{3}(x-3)$
$\Rightarrow 4x+3y=18$