TS EAMCET · Maths · Straight Lines
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 & Solution
Correct Answer
(A) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
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 \(\therefore \quad P S=P Q\) and…
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