Question

The area (in square units) of the quadrilateral formed by the point of intersection of the lines $x+y-1=0,x-y+1=0$, the point $(1,1)$ and the feet of the perpendiculars from this point on to the lines is

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

Answer 1

Answer: (A)

Equation of line are $x+y-1=0$ and $x-y+1=0$

Solving equation we get, $(0,1)$
$\therefore R(0,1)$
Perpendicular distance from $P$ to line $x+y-1=0$
and $x-y+1=0$ are $\frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{2}}$ respectively
$\therefore PS=PQ$ and $△PSR$ and
$△PQR$ are congruent triangle
$\therefore$ Area of quadrilateral $PQRS=PQ\times PS$
$=\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{1}{2}$