Question

Let $A(1,2)$ and $B(3,4)$ be two points. Let

$C(x,y)$ be a point such that$(x-1)(x-3)+(y-2)(y-4)=0$. If area of

$\left(\Delta ABC\right)=1$, then the maximum number of positions of

$C$ in $xy$-plane is

• A. $2$
• B. $4$
• C. $8$
• D. $Noneofthese$

Answer 1

Answer: (B)

$4$

$(x-1)(x-3)+(y-2)(y-4)=0$

$x^2-4x+3+y^2-6y+8=0$
$(x-2{)}^2+(y-3{)}^2-4-9+11=0$

$(x-2{)}^2+(y-3{)}^2=2$

Hence any point on the above circle can be written as

$X=Rcos\theta$ and $Y=Rsin\theta$

In the above case $R=\sqrt{2}$

Therefore

$x-2=\sqrt{2}cos\theta$

$x=2+\sqrt{2}cos\theta$...(i)

$y-3=\sqrt{2}sin\theta$
$y=3+\sqrt{2}sin\theta$ ...(ii)

Hence $C=(2+\sqrt{2}cos\theta ,3+\sqrt{2}sin\theta )$

It is given that

$A=(1,2)$
$B=(3,4)$

Hence the area of the triangle ABC using determinant method is

$=\frac{1}{2}|(3+\sqrt{2}sin\theta +4(2+\sqrt{2}cos\theta )+6)-\left(2\right(2+\sqrt{2}cos\theta )+3(3+\sqrt{2}sin\theta )+4)|$

$=\frac{1}{2}|(17+\sqrt{2}sin\theta +4\sqrt{2}cos\theta )-(17+2\sqrt{2}cos\theta +3\sqrt{2}sin\theta )|$

$=\frac{1}{2}|2\sqrt{2}cos\theta -2\sqrt{2}sin\theta |$

$=|\sqrt{2}cos\theta -\sqrt{2}sin\theta |$

$=1$

Therefore

$|cos\theta -sin\theta |=\frac{1}{\sqrt{2}}$

Dividing both sides by $\sqrt{2}$

$\left|sin(\frac{\pi }{4}-\theta )\right|=\frac{1}{2}$

$sin(\frac{\pi }{4}-\theta )=\frac{\pm 1}{2}$

Hence

$sin(\frac{\pi }{4}-\theta )=\frac{1}{2}$ implies

$\frac{\pi }{4}-\theta =\frac{\pi }{6},\frac{5\pi }{6}$

Hence

$\theta =\frac{\pi }{12},\frac{-7\pi }{12}$.

And

$sin(\frac{\pi }{4}-\theta )=\frac{-1}{2}$

$\frac{\pi }{4}-\theta =\frac{-\pi }{6},\frac{7\pi }{6}$

$\theta =\frac{5\pi }{12},\frac{-11\pi }{12}$.

Hence in total we get $4$ points.