Question

Let $P(2,-4)$ and $Q(3,1)$ be two given points. Let $R(x,y)$ be a point such that $(x-2)(x-3)+(y-1)(y+4)=0$. If area of $\Delta PQR$ is $\frac{13}{2}$, then the number of possible positions of $R$ are

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is A $2$

area of $\Delta PQR=\frac{1}{2}|\left[2\right[1-y]+3[y+4]+x[-4-1\left]\right]|$....(hint: using the determinant formula of area of triangle)

$=\frac{1}{2}|[1-2y+3y+12-xy]|$

$=\frac{1}{2}|[y-5x+14]|=\frac{\pm 13}{2}$.......(given in the question)

$\therefore 5x-y=1$ ---(1) Or $5x-y=27$ ---(2)...(taking both possibilities)

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

From 1 & 3,

$(x-2)(x-3)+(5x-2)(5x+3)=0$

$x^2-5x+6+25x^2+5x-6=0$

$26x^2=0$

$x=0$ and $y=-1$

From 2 & 3,

$x^2-5x+6+(5x-28)(5x-23)=0$

$x^2-5x+6+25x^2-255x+644=0$

$26x^2-260x+650=0$

$2x^2-20x+50=0$

$x^2-20x+50=0$

$x^2-10x+25=0$

$x=\frac{10\pm \sqrt{100-100}}{2}$

$x=5$ and $y=-2$

Therefore, there are two possible positions of R , which are $(0,-1)$ and $(5,-2)$