AP EAMCET · Maths · Straight Lines
The point on the line \(3 x+y+4=0\) which is equidistant from \((-5,6)\) and \((3,2)\) is
- A \(\left(\frac{-7}{5}, \frac{1}{5}\right)\)
- B \(\left(\frac{7}{5}, \frac{-1}{5}\right)\)
- C \((2,-2)\)
- D \((-2,2)\)
Answer & Solution
Correct Answer
(D) \((-2,2)\)
Step-by-step Solution
Detailed explanation
\( (x-(-5))^2 + (y-6)^2 = (x-3)^2 + (y-2)^2 \) \( (x+5)^2 + (y-6)^2 = (x-3)^2 + (y-2)^2 \) \( x^2+10x+25+y^2-12y+36 = x^2-6x+9+y^2-4y+4 \) \( 10x-12y+61 = -6x-4y+13 \) \( 16x - 8y + 48 = 0 \Rightarrow 2x - y + 6 = 0 \) \( (2x - y + 6) + (3x + y + 4) = 0 \)…
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