TS EAMCET · Maths · Straight Lines
\(\mathrm{A}(2,0), \mathrm{B}(0,2), \mathrm{C}(-2,0)\) are three points. Let \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) be the perpendicular distances from a variable point P on to the lines \(\mathrm{AB}, \mathrm{BC}\) and CA respectively. If \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in arithmetic progression, then the locus of P is
- A \(|\sqrt{2} y|=2|x-y+2|-|x+y-2|\)
- B \(\sqrt{2}|y|=|x-y+2|-|x+y-2|\)
- C \(2|x-y+2|=\left|\frac{x+y-2}{\sqrt{2}}\right|+\left|\frac{x-y-2}{\sqrt{2}}\right|\)
- D \(2|x-y+2|=|x+(\sqrt{2}+1) y+2|\)
Answer & Solution
Correct Answer
(A) \(|\sqrt{2} y|=2|x-y+2|-|x+y-2|\)
Step-by-step Solution
Detailed explanation
Equation of line AB: \(\frac{x}{2} + \frac{y}{2} = 1 \Rightarrow x+y-2=0\) Equation of line BC: \(\frac{x}{-2} + \frac{y}{2} = 1 \Rightarrow -x+y=2 \Rightarrow x-y+2=0\) Equation of line CA: \(y=0\) Distances from P(x, y):…
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