Question

If $\alpha$ represent the square of the distance between the origin and the point of intersection of the lines $x^{2}-y^{2}-x+3 y-2=0$ and $\beta$ represent the product of the perpendicular distances from the origin on the pair of lines, then $\alpha \beta=$

• A. $\frac{5}{4}$
• B. $1$
• C. $\frac{5}{2}$
• D. $2$

Answer 1

Answer: (C)

We have, pair of straight line is
$x^{2}-y^{2}-x+3 y-2=0$
$(x-y+1)(x+y-2)=0$
$x-y+1=0$ and $x+y-2=0$
Solving these equation, we get
$x=\frac{1}{2}$ and $y=\frac{3}{2}$
$\therefore $ Intersection point $P\left(\frac{1}{2}, \frac{3}{2}\right)$
$\alpha=O P=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{3}{2}\right)^{2}}$
$=\frac{\sqrt{10}}{2}$
Perpendicular distance from origin to line $x-y+1=0$ and $x+y-2=0$ are $\frac{1}{\sqrt{2}}$ and $\frac{2}{\sqrt{2}}$ respectively,
$\beta=\frac{1}{\sqrt{2}} \times \frac{2}{\sqrt{2}}=1$
$\Rightarrow \alpha \beta=\frac{\sqrt{10}}{2} \times 1=\sqrt{\frac{5}{2}}$